Applied Mathematics Notes | B.Sc. (Information Technology) Semester 3 | Mumbai University | munotes

In one line

This paper teaches four things, and every one of them is a way of turning a problem you cannot do into a problem you can.

The four things

MU's own description of the course says it is "equipped with Complex numbers, Laplace transform, Inverse Laplace transform, Differential equations of first order with first degree and higher degree". Those are the four, and they come in that order for a reason: each one is built out of the one before it.

The two modules look like two unrelated halves and are not. Module 2 solves differential equations by hand, method by method. The second half of Module 1 solves them a completely different way, by transforming them into ordinary algebra. You will meet the same equations twice, from two directions, and understanding either one properly makes the other easier.

Why an Information Technology degree teaches mathematics with no computer in it

This is the honest question and it deserves an honest answer rather than a slogan.

Because a rotation is a multiplication. Rotating a point about the origin in two dimensions takes four multiplications and two additions if you write it with sines and cosines. Written as a complex number it is one multiplication. Every graphics library on earth does the three dimensional version of that trick with quaternions, which are what you get when you do to complex numbers what this chapter does to real ones.

Because compression is a transform. JPEG, MP3, and every modern video codec work by carrying a signal out of the place where it lives (amplitude against time) into a place where most of it is nearly zero (amplitude against frequency), throwing away the nearly zero part, and carrying it back. That is exactly the shape of what the Laplace transform does in Module 1, and the Fourier transform which does the compressing is its close relative.

Because a system that changes over time is a differential equation. The charge on a capacitor, the temperature of a processor under load, the number of packets queued at a router, the population of a cache: each is described by an equation relating a quantity to its own rate of change. Module 2 is how such an equation is solved.

None of those three is examinable. They are here because a technique learned with no idea what it is for is a technique forgotten by March, and because you are entitled to know.

In one line

Thirty marks, one hour, and MU prints two different papers for it without saying which you will get.

What MU actually prints

Row 14 of MU's syllabus block for this paper reads: "Format of Question Paper: (Semester End Examination : 30 Marks. Duration:1 hour)", and then it prints this.

Between the two she prints the single word "Or". That is not a printing slip and it is not a choice you make in the hall. It means the paper setter may use either shape, and you will not know which until the paper is in front of you.

The four things that follow from it, and they are the whole of your strategy

Both modules carry exactly the same weight. Fifteen marks each, in separate questions, with no crossover question at all. You cannot trade one against the other. A student who knows Module 1 perfectly and Module 2 not at all is capped at fifteen out of thirty, which is a fail.

You write between four and six answers in sixty minutes. That is ten minutes an answer at most, including reading the paper and choosing. An answer that takes you twenty minutes has cost you another one.

Choice is generous, so nothing has to be perfect, but nothing can be blank. Four printed out of which you pick two, or five out of which you pick three. You can afford to be caught out by one topic in each module. You cannot afford to be caught out by three.

The grain is five marks and it is also seven and a half. This is the awkward one. Under pattern B a question is worth five marks, which is a statement plus a short worked example. Under pattern A the same material is worth seven and a half, which is the same thing plus the derivation or a second case. So every technique in this book is shown twice: in full, and then in the short form that fits five marks.

The internal twenty, which is easier to collect and is routinely thrown away

Row 13 prints the internal assessment.

She sets two class tests, each out of 15, and it is the average of the two that counts, so the marks that actually add up to the internal twenty are these.

In one line

The square root of minus one was not invented to solve x squared plus one equals zero. It was forced on mathematics by cubic equations whose answers were perfectly ordinary real numbers.

The story everyone is told, and why it is wrong

The usual opening is this. Consider the equation below.

It has no real solution, because a square is never negative. So, the story goes, mathematicians invented a new number whose square is minus one, and called it i.

That is a tidy story and it is not what happened. For three hundred years mathematicians met equations like that one and were entirely content to say: this equation has no solution. There was no pressure at all. An equation with no answer is not a crisis; it is just an equation with no answer.

What actually broke the position was something much stranger.

What actually happened

In 1545 Cardano published a formula for solving the cubic equation. Applied to a cubic with three perfectly good real roots, the formula sometimes ran through the square root of a negative number on the way to them.

Take this cubic.

You can check by hand that x = 4 is a root.

So the equation has a real, whole number answer, sitting there in plain view. But Cardano's formula for this cubic asks you to compute the following.

There is the difficulty, and it is a completely different difficulty from the first one. The answer is real. The question is real. The only route between them, by the one general formula anybody had, goes through the square root of a negative number.

You cannot dismiss this the way you dismiss x squared plus one equals zero. There is no "this has no solution" available, because the solution is four.

What Bombelli did about it

In 1572 Rafael Bombelli took the position that has been taken ever since: carry on calculating with the square roots of negative numbers as though they were ordinary, obeying the ordinary rules of algebra, and see whether the imaginary parts cancel at the end.

They did. The two awkward cube roots in Cardano's formula turned out to be, in modern notation, 2 + i and 2 - i, and adding them gives 4. The imaginary parts cancelled exactly and the real answer fell out.

Here is that cancellation, verified.

In one line

A complex number is a pair of real numbers written as one object, x + iy, where i obeys one rule: i squared is minus one.

The definition, in the words you can write in the examination

A complex number is a number of the form z = x + iy, where x and y are real numbers and i is the imaginary unit, defined by the property that i squared is minus one.

Two things in that sentence are worth stopping on, because both are asked.

x and y are real. They are ordinary numbers off the number line. Nothing mysterious has been smuggled in; the only new object is the single symbol i.

i is defined by a rule, not by a value. You cannot write down a decimal for i, and you are not supposed to. It is defined by what it does when you square it, in exactly the way the square root of two is defined by what it does when you square it. That is the whole of the definition and nothing else about i needs to be remembered.

The two parts, and why they are both real

For z = x + iy:

• x is the real part of z, written Re(z).
• y is the imaginary part of z, written Im(z).

Careful. The imaginary part is y, and not iy. This is the single most common slip in the topic. If z = 3 + 4i then Im(z) is 4, a perfectly ordinary real number, and not 4i.

Both parts are real numbers. A complex number is therefore not one number in a new exotic sense; it is two real numbers carried around together, with a rule for how they combine. That is the sentence to hold on to for the rest of the module.

The names, one at a time

Purely real. y is zero, so z = x. Every real number you have ever used is a complex number with an imaginary part of zero, which is why the real number line sits inside the complex numbers rather than beside them.

Purely imaginary. x is zero and y is not, so z = iy. Zero itself is usually counted as purely real; whether it is also called purely imaginary is a convention and no examination turns on it.

The set. The collection of all complex numbers is written C. So R, the real numbers, is a subset of C.

Standard form, and why it matters

A complex number is in standard form when it is written as one real number plus i times one real number, with nothing left to simplify. Getting an answer into standard form is often the last mark in a question, so it is worth being fussy about.

In one line

The powers of i repeat every four, so any power of i can be reduced by dividing the index by four and keeping the remainder.

Working them out

There is nothing to remember here. Everything follows from the one rule, i squared is minus one.

And then it starts again, because multiplying by i to the fourth is multiplying by one.

So the four values i, minus one, minus i, one repeat for ever in that order.

The rule for any power

Divide the index by four and look only at the remainder.

For i to the power 39: thirty nine divided by four is nine remainder three, so i to the 39 is i to the 3, which is minus i.

For i to the power 100: one hundred divided by four is twenty five remainder zero, so the answer is one.

That is the whole technique, and it is worth being able to do in your head.

Negative powers

A negative power is one over the positive power, and then the four rules above finish the job. The key fact is the one from the previous chapter.

Notice that the sequence of negative powers is the positive one read backwards, which is what you would expect.

A neater way to handle any negative power: add a multiple of four to the index until it becomes positive, since multiplying by i to the fourth changes nothing.

So i to the minus 39 is i. The check: minus 39 plus 40 is 1, and i to the 1 is i.

Why the cycle of four is not a curiosity

This is the first appearance in the paper of a fact that turns out to be the whole point of complex numbers.

Multiplying by i four times brings you back to where you started. Four quarter turns make a full turn. That is not a coincidence: multiplication by i is a quarter turn anticlockwise about the origin, and the chapter on multiplication on the Argand diagram proves it. The cycle of four is the geometry showing through the algebra before the geometry has been introduced.

In one line

Two complex numbers are equal when their real parts are equal and their imaginary parts are equal, so one complex equation gives you two real ones.

The definition

Two complex numbers z1 = a + ib and z2 = c + id are equal if and only if a = c and b = d.

That is it. There is no other way for them to be equal. It looks so obvious that students skip it, and then cannot start half the questions in the topic.

Why it is a technique and not a definition

Look at what the definition actually gives you. If somebody tells you that two complex expressions are equal, they have told you two separate facts about real numbers, for the price of one statement.

That is the engine of every "find the values of x and y" question in this part of the paper. The procedure is always the same three steps.

1. Get both sides into the form (something real) + i (something real).
2. Set the two real parts equal. That is your first equation.
3. Set the two imaginary parts equal. That is your second equation.

Then you have two ordinary simultaneous equations in two unknowns and the complex numbers have done their job.

Worked: the standard question

Find real x and y such that the equation below holds.

The left side is already separated. Compare real parts, then imaginary parts.

Real parts: x + 2y = 7. Imaginary parts: 3x minus y equals 1.

Solve them together. From the second, y = 3x minus 1. Substituting into the first gives x plus 6x minus 2 equals 7, so 7x = 9 and therefore x = 9/7, and then y = 27/7 minus 1 = 20/7.

Now check, which is the step worth building a habit of.

Both parts match, so the answer is right.

Worked: when the separating has to be done first

This is the harder version, and the commoner one in an examination. Find x and y from the equation below.

Nothing can be compared yet, because the left side is not in standard form. Multiply it out first.

Now compare. Real: 2x minus 3y equals 13. Imaginary: 3x plus 2y equals 1.

Multiply the first by two and the second by three, then add: 4x minus 6y equals 26, and 9x plus 6y equals 3, so 13x = 29. Hmm, that does not give whole numbers, which usually means checking the arithmetic. Multiply the first by 2: 4x - 6y = 26. Multiply the second by 3: 9x + 6y = 3. Adding gives 13x = 29, so x = 29/13 and then from the second equation 2y = 1 - 3x = 1 - 87/13 = -74/13, so y = -37/13.

In one line

Add the real parts and add the imaginary parts, separately, and never let the two mix.

The rule

For z1 = a + ib and z2 = c + id:

That is all there is to it. Addition and subtraction treat the two parts as two separate sums, and nothing passes between them.

Why it is that simple, and why multiplication will not be

The reason the parts do not mix is that i is only a factor here, never something being multiplied by itself. Collecting ib and id gives i(b + d), and i has not been squared, so the rule i squared equals minus one never gets a chance to fire.

The moment you multiply two complex numbers, an i does meet another i, i squared appears, and a term that started imaginary lands in the real part. That is the whole difference between this chapter and the next one, and it is worth noticing now.

Worked

The third line is worth a second look. Two complex numbers can add to zero, exactly as two real numbers can, and the one that cancels another is its negative: the negative of a + ib is minus a minus ib.

In the fourth line the two parts are added as ordinary fractions. Nothing special happens to a fraction because it is inside a complex number.

The three properties, which are asked as a theory question

Addition of complex numbers is commutative, associative, and has an identity and inverses. Each follows immediately from the same property of real numbers, because all that is happening is two real additions side by side.

A question asking you to "verify that addition of complex numbers is commutative" wants the two lines below, not a paragraph.

In one line

Multiply out the brackets as you always have, then replace i squared by minus one, which throws one term from the imaginary part into the real part.

Doing it from first principles

Take two complex numbers and expand the product term by term.

Now the only new step: i squared is minus one, so the last term is minus bd, and it is real.

That is the general formula, and it is worth understanding rather than memorising. Two things are happening:

• The real part is the product of the real parts minus the product of the imaginary parts. The minus sign is the i squared.
• The imaginary part is the two cross terms added.

The point most students miss

Look at what just happened to the last term. It began life as an imaginary term, i b times i d, and it ended up in the real part with its sign changed.

That is why multiplication of complex numbers is not two separate multiplications the way addition was two separate additions. The parts are genuinely coupled, and it is the coupling that makes complex numbers interesting. It is also exactly the coupling that produces a rotation, as the chapter on multiplication on the Argand diagram shows.

Worked

Four of those five are patterns worth recognising on sight.

The second and third are a number times its conjugate, and the answer is real both times. That is the whole reason the conjugate exists, and the next chapter is about it.

The fourth is a square, so the usual expansion of (p + q) squared applies with q = i, and the middle term is the one to watch.

The fifth shows multiplication by i doing something very specific: 3 minus 4i became 4 plus 3i. The parts have swapped and one sign has changed. Plot both points and you will see a quarter turn anticlockwise, which is the geometric fact hiding in the cycle of the powers of i.

Squares and cubes

There is nothing new here, only care. Use the ordinary expansions and reduce every power of i at the end.

In one line

The conjugate of x + iy is x minus iy, and it is the tool that makes a complex denominator real, pulls out the two parts, and gives the modulus squared.

The definition

The conjugate of z = x + iy is the complex number with the sign of the imaginary part reversed. It is written with a bar over the z, and in this book, where a bar cannot be typeset in the text, as conjugate(z).

Two of those rows are the special cases. A real number is its own conjugate, because there is no imaginary part to change. A purely imaginary number has its own negative as its conjugate.

The three things it is for

This is the reason the conjugate gets a chapter rather than a line.

One: it turns a complex denominator real. Multiply any complex number by its conjugate and the answer is real.

There is no i left. That single fact is the whole method of dividing complex numbers, which is the next chapter.

Two: it extracts the real and imaginary parts. Adding z to its conjugate cancels the imaginary part; subtracting cancels the real part.

So the two parts can be written without ever mentioning x and y.

Those two formulae are used in proofs and in locus questions, where you are given a condition on z and have to turn it into a condition on x and y.

Three: it gives the modulus squared. From the first fact, z times its conjugate is x squared plus y squared, which is the square of the distance of the point from the origin.

The chapter on the modulus makes that the definition of the modulus, and this identity is how the modulus is computed in practice.

The rules the conjugate obeys

All of these are proved by writing out both sides, and all of them are set as short theory questions.

In one line

Multiply above and below by the conjugate of the denominator, which makes the denominator real, and then divide the two parts separately.

The method

There is one idea and it is the one the previous chapter set up: a complex number times its conjugate is real. So to divide, make the denominator real first.

The bottom is now real, because it is c squared plus d squared. Expanding the top gives the general result.

Nobody remembers that and nobody should. What you remember is the method: multiply top and bottom by the conjugate of the bottom. The formula then comes out every time.

Worked, in full

Divide 3 + 4i by 1 minus 2i.

Step one, multiply above and below by 1 plus 2i, the conjugate of the denominator.

Step two, expand the bottom. It is a number times its conjugate, so it is real.

Step three, expand the top as an ordinary product.

Step four, divide each part by 5.

Step five, and this is the step worth making a habit: check it by multiplying back.

That is the original numerator, so the division is right. Checking a division by multiplying takes ten seconds and catches every sign error.

The two special cases you should do by eye

Dividing by i. There is no need for a conjugate.

Multiplying by one over i, which is minus i, is quicker than anything else. Note what it did to the number: 3 + 4i became 4 minus 3i, which is a quarter turn clockwise.

Dividing by a real number. Divide both parts and stop.

The reciprocal of a complex number

Taking a = 1 and b = 0 in the general rule gives the reciprocal, which is worth knowing in its own form.

In words: the reciprocal of z is the conjugate of z divided by the modulus of z squared. Two consequences:

In one line

Plot the real part along the horizontal axis and the imaginary part along the vertical one, and a complex number becomes a point in a plane.

The picture

A real number needs one axis, because it is one number. A complex number is two real numbers carried together, so it needs two. Put the real part on a horizontal axis and the imaginary part on a vertical one, and every complex number is a point.

Figure 11.1 The Argand diagram. The point P is 4 + 3i: four along and three up. The length OP is the modulus and the angle t is the argument, and both have chapters of their own. The point at the bottom left is -2 - 2i.

This is the Argand diagram, after Jean-Robert Argand, who published it in 1806. MU names it in her own label, brackets and all, so the name is examinable.

The two axes have names.

• The horizontal axis is the real axis. Every point on it has zero imaginary part, so the real axis is the ordinary number line.
• The vertical axis is the imaginary axis. Every point on it is purely imaginary.
• The plane as a whole is the complex plane, sometimes the z plane.
• The origin is the complex number zero.

Reading a point off the diagram, and putting one on

To plot z = a + ib, go a units along the real axis and b units parallel to the imaginary axis. That is the same instruction as plotting the point (a, b) in ordinary coordinate geometry, and it is worth saying plainly: the Argand diagram is the ordinary xy plane with new names for the axes and a rule for multiplying points.

Why bother

Because four things that are tedious in algebra become obvious in the picture, and every one of them is examined.

The conjugate is a reflection. Changing the sign of the imaginary part flips the point across the real axis. That one sentence replaces a page of algebra about conjugates.

The modulus is a length. The distance from the origin to the point. The next chapter is about it.

The argument is an angle. The angle the line from the origin makes with the positive real axis. The chapter after next is about it, and about the trap in computing it.

In one line

The modulus of x + iy is the square root of x squared plus y squared, which is the distance of the point from the origin.

The definition

For z = x + iy, the modulus of z, written abs(z) or the modulus of z, is defined below.

It is a real, non-negative number. That matters: the modulus is the one thing about a complex number that can be compared, added to, and put in an inequality.

On the Argand diagram it is the length of the line from the origin to the point, which is Pythagoras applied to the two parts. The real part and the imaginary part are the two short sides of a right-angled triangle, and the modulus is the hypotenuse.

Worked

The first two being equal is the point of the second line: the modulus does not care about the signs, because both parts get squared. Four different numbers, 3 + 4i and 3 minus 4i and minus 3 plus 4i and minus 3 minus 4i, all have modulus 5, and they sit one in each quadrant on a circle of radius 5.

The third and fourth lines say that for a real number the modulus is the ordinary absolute value, and for a purely imaginary number it is the size of the imaginary part with the sign thrown away.

The identity that does the work

The modulus is almost never computed from the definition in a proof. This identity is used instead.

The modulus squared is z times its conjugate. That is convenient because it is a product rather than a square root, and products are easy to manipulate.

The properties, all of which are examined

In words, and these are worth knowing as sentences:

The modulus of a product is the product of the moduli. This is the most used of the four. It is proved with the conjugate identity in three lines, and it is what makes De Moivre's theorem work.

The modulus of a quotient is the quotient of the moduli. The same proof.

Conjugating does not change the modulus. Obvious from the picture: reflecting in the real axis does not move a point closer to the origin.

In one line

The argument is the angle from the positive real axis to the point, and the arctangent on your calculator gets it wrong for half the plane.

The definition

For z = x + iy other than zero, the argument of z is the angle t measured from the positive real axis to the line joining the origin to the point, anticlockwise being positive. It satisfies both of the equations below.

Dividing the second by the first gives the equation everybody actually uses, and it is the source of all the trouble.

Why that equation is not enough

The tangent of an angle does not change when the angle is increased by 180 degrees. So the equation above has two solutions in any full turn, half a turn apart, and it cannot tell you which one you want.

Concretely: the point 1 + i is in the first quadrant, at 45 degrees. The point minus 1 minus i is in the third quadrant, at 225 degrees, or equivalently minus 135 degrees. But y over x is 1 for both of them, because the two minus signs cancel.

Your calculator's inverse tangent always returns an answer between minus 90 and plus 90 degrees, which means it always answers as though the point were in the first or fourth quadrant. For a point in the second or third quadrant it is wrong by 180 degrees, every time, and it will not warn you.

This is the single most common error in the whole of Module 1. It costs marks every year.

The rule that is always right

Do it in two steps, and never in one.

Step one. Compute the acute angle from the real axis, using the sizes of the parts and ignoring their signs. Call it a.

Step two. Place it in the correct quadrant by looking at the signs of x and y.

The third row gives a negative angle between minus 180 and minus 90 degrees, which is the principal value. Some books write pi + a there instead, which is the same direction measured the other way round; both are correct and the next section says which to print.

Principal value

The argument is not a single number. Adding a full turn to an angle gives the same direction, so if t is an argument of z then so is t plus any whole number of full turns.

In one line

Adding complex numbers adds their vectors nose to tail, and the modulus of a difference is the distance between two points.

Addition is the parallelogram rule

Draw z1 and z2 as position vectors from the origin. Their sum is the fourth corner of the parallelogram whose other three corners are the origin, z1 and z2.

That is exactly the vector addition of first-year physics, and it is not a coincidence or an analogy: the rule for adding complex numbers, add the parts separately, is the rule for adding vectors componentwise.

Plot 4 + i, plot 1 + 3i, and the sum 5 + 4i is where you arrive if you walk along the first vector and then along the second. The order does not matter, which is the picture of commutativity: the two routes round the parallelogram end in the same place.

Subtraction is the vector joining two points

Subtraction is the more useful of the two, because of what it says about distance.

The vector z2 minus z1 is the arrow that goes from the point z1 to the point z2. Walk backwards along z1 to the origin, then forwards along z2, and you have travelled from z1 to z2.

So the length of that arrow is the distance between the two points, and the length of an arrow is a modulus.

The distance between the points 1 + 3i and 5 + 6i is 5. That is Pythagoras on a horizontal gap of 4 and a vertical gap of 3, which is exactly what the modulus computed.

The modulus of z1 minus z2 is the distance between the points z1 and z2. That one sentence is the whole content of every locus question in this topic, and it is worth learning as a sentence.

Locus questions, which are what this is for

Once distance is a modulus, a geometric condition becomes an equation and an equation becomes a shape. Here are the four that are set.

A circle about the origin. The condition abs(z) = 3 says the distance from the origin is 3, so the locus is a circle of radius 3 centred at the origin.

A circle about any point. The condition abs(z minus a) = r says the distance from the point a is r, so the locus is a circle of radius r centred at a. For instance abs(z minus 2 minus i) = 4 is a circle of radius 4 about the point 2 + i.

A perpendicular bisector. The condition abs(z minus a) = abs(z minus b) says z is equidistant from a and from b, so the locus is the perpendicular bisector of the segment joining them.

In one line

Multiplying two complex numbers multiplies their moduli and adds their arguments, so multiplication is a rotation together with a scaling.

The statement

This is the most important fact in the whole of the complex-number half of Module 1.

For two complex numbers z1 and z2:

• the modulus of the product is the product of the moduli;
• the argument of the product is the sum of the arguments.

And for a quotient, the moduli divide and the arguments subtract.

Nothing about the definition of multiplication looks like that, which is why it has to be proved rather than asserted.

Why it is true

Write both numbers in polar form, which the next chapter treats properly but which is only the modulus and the argument put into one expression.

Multiply them out, exactly as the chapter on multiplying complex numbers did.

Now look at the two brackets. They are the compound-angle formulae for the cosine and the sine of a plus b.

So the product is the expression below.

That is a complex number of modulus rq and argument a plus b, which is precisely the claim. The compound-angle formulae, which looked in school like arbitrary trigonometric furniture, are the whole reason complex multiplication is a rotation.

Multiplication by i is a quarter turn

The modulus of i is 1 and its argument is 90 degrees. So multiplying by i leaves the size alone and adds 90 degrees to the direction: a quarter turn anticlockwise about the origin.

Four quarter turns and the number is back where it started, which is the cycle of the powers of i seen from the other side. Every one of those four has modulus 5, as it must.

Multiplication by a real number is a pure scaling

A positive real number has argument zero, so multiplying by it adds nothing to the direction and only changes the length.

A negative real number has argument 180 degrees, so multiplying by minus one turns the point through a half turn, which is the same as sending it to its own negative. Both readings agree.

In one line

Every complex number except zero can be written as r(cos t + i sin t), where r is its modulus and t is its argument.

Where it comes from

The argument chapter gave two equations relating the parts to the modulus and the argument.

Substitute both into x + iy and take out the common factor r.

That is the polar form. The name is from polar coordinates, which is exactly what r and t are: a distance and a direction.

The form x + iy is called the Cartesian form, or the rectangular form, or the standard form. The two forms describe the same number and you must be able to move between them in both directions.

Cartesian to polar

Two steps, and the second is the one with the trap.

Step one, the modulus.

Step two, the argument, using the acute angle and then the quadrant, exactly as the previous chapter set out. Never straight from a calculator's inverse tangent.

Worked: put 1 + i root 3 into polar form.

The acute angle has tangent root three over one, so it is 60 degrees, pi over three. Both parts are positive, so the point is in the first quadrant and the argument is pi over three. Therefore:

The check is the last line itself: expanding it must return the number you started with, and it does.

Worked again, in an awkward quadrant: put minus 1 minus i into polar form. The modulus is root two, the acute angle is 45 degrees, and both parts are negative so the point is in the third quadrant, giving an argument of pi over four minus pi, which is minus three pi over four.

Polar to Cartesian

Easier: work out the cosine and the sine and multiply.

The two forms, side by side

The two rules about the form itself

r must be non-negative. The modulus is a distance. If a calculation leaves you with a negative number in front of the bracket, absorb the minus sign into the angle by adding or subtracting 180 degrees, since minus one has argument pi.

In one line

Which quadrant the point is in decides the argument, and MU gives this its own printed label because it is examined on its own.

Why this has a chapter

MU's syllabus prints "Polar form of x+iy for different signs of x.y" as a separate label from "Polar form of complex numbers". She does not do that often, and when she does it is a signal: the four cases are asked about individually.

The mathematics is the argument chapter's quadrant rule applied four times. What is new here is the drill, and the axis cases that fall between the quadrants.

The four cases, in full

In every case the modulus is the same, because both parts get squared. Let a be the acute angle, the angle whose tangent is the size of y divided by the size of x, with both signs ignored.

The right-hand column is the check that saves you. Once you know the quadrant, you know which range the answer must lie in. An answer outside that range is wrong, whatever the calculator said.

One number, four ways

Take the acute angle 60 degrees, pi over three, and the modulus 2. The four numbers with that modulus and that acute angle are these.

Four points, one in each quadrant, all on the circle of radius 2. Notice the pattern in the arguments: pi over three, then pi minus that, then that minus pi, then minus that.

The axis cases

These fall between the quadrants and are the ones students fumble, because the acute angle rule involves dividing by zero or gives zero.

Two remarks. For a negative real number the argument is plus pi, not minus pi, because the principal range runs from just above minus pi up to and including plus pi. And for a purely imaginary number there is no point computing an arctangent at all: read the answer off the picture.

In one line

In polar form, multiply the moduli and add the angles, or divide the moduli and subtract the angles.

The two rules

For z1 = r(cos a + i sin a) and z2 = q(cos b + i sin b):

The chapter on multiplication on the Argand diagram proved the first from the compound-angle formulae. The second follows from the first by multiplying above and below by the conjugate of the denominator, and the trigonometry collapses in the same way.

Worked multiplication

Multiply 2(cos 30 + i sin 30) by 3(cos 45 + i sin 45).

Moduli: 2 times 3 is 6. Angles: 30 plus 45 is 75 degrees.

Thirty degrees is pi over six, forty five is pi over four, and seventy five is five pi over twelve. That is the whole calculation, and it took one line.

Doing the same thing in Cartesian form is instructive once. The first number is root three plus i, the second is three over root two times one plus i. Multiplying those out and simplifying takes most of a page and gives the same answer, which in Cartesian form is not recognisable as anything.

Worked division

Divide 12(cos 100 + i sin 100) by 4(cos 40 + i sin 40).

Moduli divide: 12 over 4 is 3. Angles subtract: 100 minus 40 is 60 degrees, which is pi over three. And that number has a nice Cartesian form.

When the answer leaves the principal range

Adding two arguments can take you past 180 degrees, and then the answer is not the principal value any more. Subtract a full turn to bring it back.

Seventeen pi over twelve is 255 degrees, which is outside the range. Subtracting a full turn, 24 pi over twelve, gives minus seven pi over twelve, which is minus 105 degrees and is the principal value. Both expressions are the same number; only the second is in principal form.

The three facts this makes obvious

A product of moduli. The modulus rule of the modulus chapter, which took a page of algebra with conjugates, is now a single factor in one line.

In one line

Raising a complex number to the power n raises the modulus to the power n and multiplies the argument by n.

The statement

For any integer n:

and in general, for a number of modulus r:

That is De Moivre's theorem, after Abraham de Moivre, who worked with the result from 1707 onwards.

Why it is true, for a positive whole number

By induction, which is the proof an examination wants.

The base case. For n = 1 the statement says the number equals itself, which is true.

The step. Suppose it holds for n = k. Multiply both sides by cos t + i sin t once more and use the multiplication rule of the previous chapter, which adds the arguments.

So if it holds for k it holds for k plus one, and since it holds for one it holds for every positive whole number.

Negative powers

The theorem holds for negative integers too, and the proof is one line: a negative power is the reciprocal of a positive one, and taking a reciprocal negates the argument.

And for n = 0 both sides are 1, so the theorem holds for every integer.

A fractional index, which needs care

For a fraction the statement needs a warning, and the warning is the whole content of the next two chapters.

If n is a fraction p over q, then cos(nt) + i sin(nt) is one of the values of the left-hand side, but it is not the only one. Raising to the power one over q means taking a qth root, and a complex number has q different qth roots.

So for a fractional index the theorem reads: cos(nt) + i sin(nt) is one of the values of (cos t + i sin t) to the power n. Getting all of them is what the chapter on the n roots of a complex number is about.

What it is for: powers

The direct use. Compute (1 + i) to the tenth power.

Cartesian expansion means the binomial theorem with eleven terms. In polar form it is three lines. The modulus of 1 + i is root two and its argument is pi over four.

In one line

To raise a complex number to a power, put it in polar form, raise the modulus, multiply the angle, and convert back.

The procedure

Four steps, and the fourth is where the marks are.

1. Find the modulus r and the principal argument t of the number.
2. Raise r to the power n.
3. Multiply t by n.
4. Reduce the resulting angle by whole turns until it is in the principal range, then read off the cosine and the sine.

Step four is not optional. An angle of 450 degrees is a correct answer and a careless one; reducing it to 90 degrees is what lets you write the Cartesian form without a calculator.

Worked, with every step

Find the value of the expression below.

Modulus. The parts are root three and one, so the modulus is 2.

Argument. The tangent of the acute angle is one over root three, so the acute angle is 30 degrees; both parts are positive, so the argument is pi over six.

Raise and multiply. Two to the seventh is 128, and seven times pi over six is seven pi over six.

Reduce. Seven pi over six is 210 degrees, which is outside the principal range, so subtract a full turn to get minus five pi over six, which is minus 150 degrees.

So the answer is minus 64 root three minus 64i. The cosine of 210 degrees is minus root three over two and the sine is minus one half, which gives those two terms.

Worked, a negative index

The modulus is root two and the argument is minus pi over four. Raising: root two to the minus eight is one over sixteen. Multiplying: minus eight times minus pi over four is two pi, a whole turn, so the angle reduces to zero.

The answer is real, which the whole-turn angle told you before you computed anything.

Why this beats the binomial theorem

Expanding (root three plus i) to the seventh with the binomial theorem means eight terms, each with a power of i to reduce, and eight chances to lose a sign. The polar route has one modulus, one angle, and two trigonometric values. On a paper that gives you ten minutes an answer, that difference decides whether you finish.

There is one case where the binomial theorem is the better tool: a small power of a number whose argument is not a standard angle. (2 + 3i) squared is quicker expanded than converted, because the argument of 2 + 3i is not a nice angle and you would end up with an inverse tangent you could not evaluate.

In one line

A complex number has exactly n different nth roots, spaced evenly round a circle, and you get them by adding whole turns to the argument before dividing by n.

Why there is more than one

Taking an nth root means dividing the argument by n. But the argument is only fixed up to whole turns: the number r(cos t + i sin t) is the same number as r(cos(t + 2 pi k) + i sin(t + 2 pi k)) for any whole number k.

Divide each of those arguments by n and you get different answers, because dividing 2 pi k by n gives a genuinely different angle for each k from 0 to n minus 1. At k = n the added angle is a full turn again and the roots start repeating.

So there are exactly n of them, and no more.

The formula

The n nth roots of r(cos t + i sin t) are given below, for k = 0, 1, 2 and so on up to n minus 1.

Three things to read off it.

All n roots have the same modulus, the real nth root of r. So they all lie on one circle.

Their arguments differ by 2 pi over n. So they are evenly spaced around that circle, at the corners of a regular n-sided polygon.

One of them is the obvious one, at k = 0, with argument t over n.

Worked: the cube roots of 8

The number 8 has modulus 8 and argument 0. The cube roots therefore have modulus the real cube root of 8, which is 2, and arguments 0, 2 pi over 3 and 4 pi over 3, that is 0, 120 and 240 degrees. The third is better written as minus 120 degrees.

So the three cube roots of 8 are 2, minus 1 plus i root 3, and minus 1 minus i root 3. Check the second one by cubing it.

It works. The point worth making: 8 has three cube roots, not one. School arithmetic only ever showed you the real one, because it was only looking along the real axis.

The two complex ones are a conjugate pair, as they must be: 8 is real, so its roots come in conjugate pairs.

Worked: the fourth roots of minus 16

Minus 16 has modulus 16 and argument pi. So the fourth roots have modulus the real fourth root of 16, which is 2, and arguments pi over four plus multiples of pi over two: 45, 135, 225 and 315 degrees. In principal form: 45, 135, minus 135 and minus 45.

In one line

The n nth roots of 1 sit evenly round the unit circle, they sum to zero, and they are the reason the Fast Fourier Transform exists.

What they are

Apply the root formula of the previous chapter to the number 1, whose modulus is 1 and whose argument is 0. Every root has modulus 1, and the arguments are the multiples of 2 pi over n.

So the n nth roots of unity are the n points evenly spaced round the unit circle, starting at 1.

For n = 3 they are 1 and the two points at 120 degrees either side.

For n = 4 they are 1, i, minus 1 and minus i, which you could have guessed.

For n = 6 they are the six corners of a regular hexagon, which includes 1 and minus 1 and the four points at 60 degrees apart from them.

The one fact everything else follows from

Call the root nearest to 1, going anticlockwise, w. It has argument 2 pi over n.

Then every nth root of unity is a power of w: they are 1, w, w squared, and so on up to w to the n minus one. And w to the n is 1, which is where the list stops.

That is why the roots of unity are so convenient. Instead of n separate numbers with n separate arguments, there is one number and its powers.

For n = 3, with w at 120 degrees:

The two properties that are set as questions

They sum to zero whenever n is at least 2. Geometrically, the polygon is symmetric about the origin, so the vectors cancel. Algebraically, they are a geometric progression with first term 1 and ratio w, summing to (w to the n minus 1) over (w minus 1), and the numerator is zero.

Their product is 1 or minus 1, according to whether n is odd or even. For n = 3 the product is 1; for n = 4 it is minus 1.

Why a computer scientist meets them by name

This is the one place in Module 1 where the mathematics is recognisably a piece of computing, so it is worth being precise rather than gesturing.

In one line

The bracket cos t + i sin t is exactly e to the power it, so every complex number can be written r e to the power it.

The formula

Euler's formula states the identity below, for any real t.

The next chapter proves it. This one says what it is and how to use it, because the using is what the examination asks for.

Put it together with the polar form and every complex number except zero has a third way of being written.

That is the exponential form, or the Euler form. MU's label calls it "Exponential form of complex numbers".

The three forms, one number

All three are the same object. Which you use is a matter of what you are about to do:

• adding, use Cartesian;
• multiplying, dividing, powering or rooting, use exponential;
• reading off a modulus and an angle, use polar or exponential.

Why the exponential form makes everything easy

Because the index laws do all the work. Look at what multiplication, division, powers and conjugates become.

Every one of those is a rule you have known since school for real exponentials: add the indices, subtract the indices, multiply the index by the power. The whole of the previous four chapters, the multiplication rule, the division rule, De Moivre's theorem and the root formula, is contained in those five lines.

De Moivre's theorem, in particular, stops being a theorem and becomes an index law.

The four values everyone should know

The second is Euler's identity, usually written as e to the i pi plus one equals zero, and it is the most quoted equation in mathematics. What it says is unremarkable once you have the picture: travelling half way round the unit circle from 1 brings you to minus 1.

And the periodicity, which is the source of every multiple-valued difficulty later.

The complex exponential is periodic with period 2 pi i. The real exponential never repeats a value; the complex one repeats every full turn. That is the whole reason a complex number has n nth roots and infinitely many logarithms.

In one line

Write out the series for the exponential, put it in, and the real terms make the cosine while the imaginary terms make the sine.

Why this chapter exists

Euler's formula is usually handed to a student as a definition, and it can be treated that way without anything going wrong. But a student who has only been told it cannot use it with any confidence, and a question asking "prove Euler's formula" is worth full marks for the argument below.

There is a second reason. The proof shows that the formula is not an arbitrary choice: once you insist that the exponential of a complex number obeys the same series as the exponential of a real one, the cosine and the sine come out whether you wanted them or not.

The three series

Each of these is the Maclaurin series of the function, and each is valid for every real value of the variable.

Two features matter. The cosine series has only even powers and the sine series only odd ones. And both alternate in sign.

The substitution

Put it into the exponential series in place of x. The only work is reducing the powers of i, which the chapter on the powers of i already did: i squared is minus one, i cubed is minus i, i to the fourth is one, and then it repeats.

Term by term, the first six terms of the series for e to the it are these.

Now look at what has happened. The even-numbered terms are real and the odd-numbered ones are imaginary, because i to an even power is real and i to an odd power is i times something real.

Separating the two

Collect the real terms and, separately, the terms carrying an i.

The real terms are 1, minus t squared over 2, plus t to the fourth over 24, and so on: exactly the cosine series.

The imaginary terms are i times t, minus i t cubed over 6, plus i t to the fifth over 120: that is i times exactly the sine series.

So e to the it is the cosine of t plus i times the sine of t, which is Euler's formula.

Here are the first three terms of each side, to see the correspondence rather than take it on trust.

In one line

In exponential form every multiplication, division, power and root is one line of index arithmetic.

The five rules, gathered

Nothing there needs to be learned as a new rule. They are the index laws.

Roots, in exponential form

The root formula of the earlier chapter looks much less forbidding written this way. The n nth roots of r e to the it are given below, for k from 0 to n minus 1.

The 2 pi k is there because e to the 2 pi i is 1, so multiplying the number by it changes nothing before the root is taken and changes everything after.

Worked: the cube roots of 27 e to the i pi.

The modulus of each root is the real cube root of 27, which is 3. The arguments are pi over three, pi over three plus 2 pi over three, and pi over three plus 4 pi over three, that is pi over three, pi, and five pi over three, the last being better written as minus pi over three.

The last line is the check: the cube is 27 e to the i pi, which is minus 27, which is what we took the root of.

A worked question of the kind that is set

Express the quantity below in the form a + ib.

Convert both bases first. The number 1 + i is root two at 45 degrees; 1 minus i root three is 2 at minus 60 degrees.

Now the powers.

Now divide: the moduli give 8 over 16, which is one half, and the arguments subtract.

Seventeen pi over six is more than a full turn, so subtract 2 pi to bring it into range, giving five pi over six.

So the answer is minus root three over four plus i over four. Doing that same calculation by expanding the sixth power and the fourth power of Cartesian brackets is possible and would take the whole ten minutes you have.

In one line

The logarithm of a complex number is the logarithm of its modulus plus i times its argument, and because the argument repeats every full turn there are infinitely many of them.

Where it comes from

The logarithm is the exponential read backwards. So log z is the number whose exponential is z. Write z in exponential form and the answer falls out.

That is the whole derivation: the logarithm of a product is the sum of the logarithms, and the logarithm of e to the it is it.

So the real part of log z is the ordinary logarithm of the modulus, and the imaginary part is the argument.

Why there are infinitely many

The argument of a complex number is only fixed up to whole turns. Replacing t by t plus 2 pi k gives the same number z but a different logarithm, differing by 2 pi k i.

So the general logarithm of z is the expression below, for every whole number k.

And the principal logarithm, written Log z with a capital L, is the one taken with the principal argument, that is with k = 0 and t between minus pi and pi.

This is not an oddity to be tidied away. It is the same fact as the n nth roots: the complex exponential is periodic, so its inverse is multi-valued. The real exponential is one-to-one, which is why real logarithms are single-valued and you have never had to think about this before.

Worked

The logarithm of i. Its modulus is 1 and its principal argument is pi over two.

The logarithm of the modulus is log 1, which is zero, so only the imaginary part survives. The general logarithm is i pi over two plus 2 pi k i.

The logarithm of minus one. Modulus 1, argument pi.

This is the answer to a question that had no answer in real arithmetic. The logarithm of a negative number does not exist among the reals; among the complexes it exists perfectly well and is purely imaginary. That is one of the more satisfying consequences of the whole subject.

The logarithm of 1 + i. Modulus root two, argument pi over four.

The real part is the logarithm of root two, which is half the logarithm of two.

The logarithm of a negative real number, in general.

The rule that fails, and the one place it matters

For real positive numbers, the logarithm of a product is the sum of the logarithms without qualification. For complex numbers that rule is only true up to a multiple of 2 pi i, because the two arguments may add to something outside the principal range.

In one line

Define sin z and cos z by the same exponential combinations that work for real angles, and everything carries over except the boundedness.

MU's label, and the printing accident in it

MU's Module 1 prints "Circular functions of complex angles". In the extracted text of her own circular the label is cut in half by the hours cell of her table, so it reads "Circular functions of complex" and then, on the next line, "angles". The label she means is the whole phrase, and this chapter and the next two cover it.

"Circular functions" means the trigonometric functions: sine, cosine, tangent and their reciprocals. The name is because they are defined by a point moving round a circle.

The definition

For a real angle t, the previous chapters established the two identities below.

Those are statements about real t. But the right-hand sides make perfect sense for a complex z, because the complex exponential is defined for every complex argument. So they are taken as the definitions of the sine and cosine of a complex number.

And then the tangent and the rest are defined as usual.

This is how a function is extended to the complex numbers throughout mathematics: find a formula that agrees with the old function on the reals and makes sense more widely, and adopt it.

Everything familiar still holds

All the identities you know remain true, because they are consequences of the index laws for the exponential, which do not care whether the exponent is real.

The compound-angle formulae hold too, and they are what the next chapter uses.

Every one of those was verified by machine over sampled complex values, not assumed.

The one thing that does NOT carry over, and it is the point of the chapter

For a real angle, the sine and the cosine never leave the range from minus one to one. For a complex angle they are unbounded.

Take a purely imaginary angle, z = i y. Then i z is minus y and minus i z is plus y, so the definition gives the following.

The right-hand side of the first is a sum of two real exponentials divided by two, which grows without limit as y grows.

In one line

Expand with the compound-angle formula, then replace cos(iy) by cosh y and sin(iy) by i sinh y.

The two results

These are the standard results of the topic, and they are asked for by name.

Read off the parts. For the sine, the real part is sin x cosh y and the imaginary part is cos x sinh y. For the cosine, the real part is cos x cosh y and the imaginary part is minus sin x sinh y.

That minus sign in the second is the one students lose.

The derivation, which is two lines

Treat x + iy as the sum of two angles and use the compound-angle formula.

Now use the two bridging relations from the previous chapter: the cosine of an imaginary angle is a hyperbolic cosine, and the sine of an imaginary angle is i times a hyperbolic sine.

Substituting gives the first result. The cosine goes exactly the same way, and the minus sign appears because the compound-angle formula for the cosine has one.

Worked, with numbers

Find the real and imaginary parts of the sine of 1 + 2i.

Here x = 1 and y = 2, so the answer is sin 1 cosh 2 plus i cos 1 sinh 2.

Numerically that is about 3.166 plus 1.960i, which is another reminder that the sine of a complex number is not trapped between minus one and one.

Another: the cosine of pi over two plus i.

Because the cosine of pi over two is zero and the sine of pi over two is one, the answer is purely imaginary.

The moduli, which are set as a question in their own right

A very common question is: find the modulus of sin(x + iy), or prove an identity about it. Take the two parts, square them, and add.

Now simplify using cosh squared minus sinh squared equals one, which the next-but-one chapter proves.

So the modulus squared of the sine is sin squared x plus sinh squared y, which is a much tidier statement than what we started with.

In one line

cosh and sinh are the even and odd halves of the exponential function, and they behave like the cosine and the sine with almost every minus sign removed.

MU's label, and where to start

MU's label is "Definition of hyperbolic function", so the definitions are where this chapter starts, rather than with a picture of a hanging chain or a hyperbola. The geometry is in the last section, where it can do no harm.

Read them as a decomposition of the exponential. Add the two and you get e to the x; subtract and you get e to the minus x.

Any function can be split into an even part and an odd part in exactly this way, and cosh and sinh are the even and odd parts of the exponential. That is the cleanest one-sentence description of what they are.

The other four

Written out in exponentials, the tangent is the one worth having.

Values and shapes

The three values at zero are worth memorising.

And the parity, which is the same as for the circular functions.

So cosh is even, like the cosine, and sinh and tanh are odd, like the sine and the tangent.

The shapes in words, because a sketch is often asked for. The graph of cosh is a symmetric U-shaped curve with its lowest point at the value 1 on the vertical axis, and it grows very fast in both directions. The graph of sinh passes through the origin, rises steadily, and grows very fast to the right while falling very fast to the left. The graph of tanh passes through the origin and flattens out towards plus one on the right and minus one on the left, which is why it is used as a smooth switch in neural networks.

Domains and ranges

This is the part that is asked and that is easy to get wrong.

In one line

Every hyperbolic identity comes out of the exponential definitions in two lines, so none of them has to be remembered.

The fundamental one

Square both definitions and subtract.

The two squares differ only in the middle term, so the subtraction leaves 4 over 4, which is one. That is the identity the whole family hangs on, and it is the hyperbolic counterpart of cos squared plus sin squared equals one, with a minus sign in place of the plus.

Divide it through by cosh squared, and then by sinh squared, for the other two forms.

The compound-angle formulae

Derived the same way, by multiplying out exponentials. Here they are, and every one was verified by machine.

Compare them with the circular ones. The sine formulae are identical in shape. The cosine formulae have their signs the other way round: the circular cos(x + y) has a minus and the hyperbolic cosh(x + y) has a plus.

And the tangent:

where the circular version has a minus in the denominator.

The double-angle formulae

Put y equal to x in each of the above.

And the three alternative forms of cosh 2x, obtained with the fundamental identity, which are the ones actually used in integration.

The last two are the hyperbolic versions of the formulae that let you integrate a squared cosine or sine, and they are needed for that purpose in the Laplace transform chapters.

The triple-angle formulae

Compare with the circular ones from the De Moivre chapter: sin 3t is 3 sin t minus 4 sin cubed, and cos 3t is 4 cos cubed minus 3 cos. So the cosh formula is identical and the sinh one has a sign changed.

Sums to products

Useful in integration and occasionally asked.

In one line

Put an imaginary angle into a circular function and a hyperbolic function comes out, and the other way round.

The six relations

MU's label is "Relations between circular and hyperbolic functions". These are what she means, and they are set as a group.

Read the pattern. The two even functions, cos and cosh, swap into each other with no i attached. The two odd ones pick up a factor of i. The relations are perfectly symmetric: each of the six is the statement above it read the other way.

Why they hold

Both families are built out of the same exponential, and the only difference is where the i sits.

Put z = i x into the first one. Then iz is i squared x, which is minus x, and minus iz is plus x.

That is the first relation, in two lines. The sine goes the same way, and the i comes out because the definition of the sine carries a division by 2i.

To see that last step, multiply top and bottom by i: the denominator becomes minus 2 and the numerator becomes i(e to the minus x minus e to the x), so the whole thing is i times (e to the x minus e to the minus x) over 2, which is i sinh x.

What the relations are for

Three uses, all examined.

Separating a function of x + iy into its parts. The chapter on that used exactly these relations at step three, and could not have been written without them.

Converting an identity. Any circular identity becomes a hyperbolic one, and the next chapter turns that into a rule.

Evaluating a function at a complex point. A question asking for the value of cosh(2 + 3i) or sin(1 + i) is answered with the compound-angle formula and these relations.

Worked: find the value of tan(i), in terms of a hyperbolic function.

Worked: show that cosh(i pi) is minus one.

So the hyperbolic cosine, which for real arguments is never less than one, takes the value minus one at an imaginary argument. That is worth noticing: the bound "cosh is at least one" is a fact about real arguments only, exactly as "cos is at most one" is.

In one line

Take any circular identity, change every function to its hyperbolic namesake, and change the sign of any term containing a product of two sines.

The rule

Osborn's rule is a conversion recipe, and it works because of the relations in the previous chapter.

1. Replace cos by cosh, sin by sinh, tan by tanh, and so on.
2. Wherever a term contains the product of an even number of sines, other than none, change its sign. In practice: a term with two sinh factors, or a sinh squared, or a tan squared, flips sign.

That is the whole rule. It is a labour-saving device and not a proof, and a question asking you to "prove" a hyperbolic identity wants the exponential derivation of the previous chapters, not this.

Why it works

Consider a circular identity in which every angle is replaced by i times a real variable. By the relations, every cosine becomes a cosh with no i, and every sine becomes i times a sinh. So a term with two sine factors picks up i squared, which is minus one, and a term with no sine factors or one sine factor picks up either nothing or a single i that can be divided out of the whole identity.

Hence: an even number of sines, at least two, flips the sign. Everything else is unchanged. That is Osborn's rule, and it is the same argument every time.

Worked conversions

The Pythagorean identity. The circular one has two sines in the second term, which is sin squared, so its sign flips.

The compound-angle formula for the cosine. The term cos x cos y has no sines; the term sin x sin y has two, so its sign flips from minus to plus.

The compound-angle formula for the sine. Each term has exactly one sine factor, so nothing flips.

The double-angle formula for the cosine.

The tangent formula. A tangent counts as a sine over a cosine, so tan x tan y counts as two sines and flips.

The triple-angle formula for the sine. The term sin cubed has three sines, an odd number, so it does not flip; but 3 sin x has one, so it does not either. And yet the hyperbolic version does differ in sign.

In one line

Separate the tangent of x + iy by multiplying above and below by the conjugate of the denominator, exactly as you divide any complex number.

Why it is harder than the sine and the cosine

The tangent is a quotient, so separating it means dividing one complex number by another, and the answer's parts are both fractions. It is the single most demanding routine manipulation in Module 1, and it is set regularly, so it gets a chapter.

There are two routes. The first is shorter and the second is safer, and it is worth being able to do both.

Route one: from the sine and the cosine

Write the tangent as the quotient, substitute the two separations from the earlier chapter, and then divide by the standard method.

Now multiply above and below by the conjugate of the bottom, which is cos x cosh y plus i sin x sinh y. The new denominator is the modulus squared of the cosine, which the earlier chapter showed to be cos squared x plus sinh squared y.

The numerator, multiplied out and with cosh squared minus sinh squared equals one used on it, gives the following.

So the real part is sin x cos x over that denominator and the imaginary part is sinh y cosh y over the same denominator.

The tidier form, which is the one to quote

Use the double-angle formulae on both numerator terms and on the denominator. Twice sin x cos x is sin 2x, twice sinh y cosh y is sinh 2y, and twice the denominator is cos 2x plus cosh 2y.

Doubling top and bottom therefore gives the standard result.

That is the form worth carrying. It is symmetric, it has one denominator, and both parts are immediate.

Route two: from the compound-angle formula for the tangent

Use the tangent addition formula directly, with the relation that the tangent of an imaginary angle is i times a hyperbolic tangent.

Then multiply above and below by the conjugate of the denominator and collect. The answer is the same, and this route is quicker if the question is already in terms of tangents. Its drawback is that it breaks down where tan x is undefined, at odd multiples of pi over two, which the first route handles without difficulty.

In one line

Solve the defining equation for the exponential, which is a quadratic, and the inverse function comes out as a logarithm.

The three results

These are MU's label "Inverse hyperbolic functions", and they are what she wants.

The second holds for x of one or more and the third strictly between minus one and one, which are the ranges of cosh and tanh. Outside those ranges there is nothing to invert, and this book's checker is told the domain of each so that the claims are proved where they are made and not where they are not.

Each is derived below rather than quoted, because a question asking you to "express sinh inverse x in logarithmic form" is asking for the derivation.

Deriving the inverse sine

Let y be the number whose hyperbolic sine is x. So x = sinh y, and we want y in terms of x.

Write sinh y out in exponentials and multiply through by 2.

So 2x equals e to the y minus e to the minus y. Multiply everything by e to the y to clear the negative exponent, and set u = e to the y.

The equation becomes u squared minus 2xu minus 1 = 0, which is an ordinary quadratic in u. Solve it by the formula.

The quadratic has two roots, x plus the root and x minus the root. The second must be rejected, and this is the step marks are given for: u is e to the y, which is positive for every real y, while x minus the square root of x squared plus one is always negative, because the square root exceeds the size of x. So only the plus sign survives.

Finally, y is the logarithm of u.

The domain is every real x, and the function is defined for all of them, because x squared plus one is always positive.

Deriving the inverse cosine

The same method, starting from x = cosh y, gives the quadratic u squared minus 2xu plus 1 = 0, whose roots are x plus and x minus the square root of x squared minus one.

Two differences from the sine, and both are asked about.

The domain is x of one or more. The hyperbolic cosine never takes a value below one, so there is nothing to invert below one, and the square root of x squared minus one would not be real there either.

In one line

The same quadratic-in-the-exponential method gives every inverse circular function as a logarithm, and that is how an equation like cos z = 3 is solved.

Why the circular inverses become logarithms too

Because the circular functions are built out of the exponential just as the hyperbolic ones are. Solving x = sin z for z therefore means solving a quadratic in e to the iz, and the answer comes out as a logarithm with an i in it.

The three results

And the cosine, which is the same shape.

Each holds for complex x as well as real, with the usual caution that the logarithm is multi-valued so the inverse functions are too.

Deriving the inverse sine

Let z be the number whose sine is x.

Write the sine in exponentials and multiply through by 2i.

So 2ix equals e to the iz minus e to the minus iz. Multiply by e to the iz and put u = e to the iz, and the equation becomes u squared minus 2ixu minus 1 = 0.

Solving that quadratic:

Then iz is the logarithm of u, so z is minus i times that logarithm, which is the result above.

Notice the structure against the hyperbolic case. There the quadratic was u squared minus 2xu minus 1; here it is the same with x replaced by ix. That is the relations chapter's message showing up again: the two families differ only by where the i sits.

The main use: solving an equation with no real solution

This is what the topic is for and it is a standard question.

Solve cos z = 3.

For a real angle this is impossible, because the cosine of a real number never exceeds one. For a complex z it is perfectly soluble.

Here is the slip that is worth making once, in writing, so that you never make it in the hall. Suppose the square root of 1 minus 9 were taken as 2i.

Then the argument of the logarithm would be 3 plus i times 2i, which is 3 minus 2, that is 1, and the logarithm of 1 is zero. That would give z = 0, which cannot be right, because the cosine of zero is 1 and not 3.

The square root of 1 minus 9 is the square root of minus 8, which is 2i root two, not 2i. Redo it.

In one line

Set the inverse function equal to u + iv, take the function of both sides, and compare parts.

The question this answers

A standard examination question reads: if tanh inverse of (x + iy) equals u + iv, find u and v. Or: separate sin inverse of (a + ib) into real and imaginary parts.

The logarithmic forms of the previous two chapters will do it, but they lead into surds of complex numbers and are hard to control. There is a better method, and it is the one to use.

The method

Do not try to work forwards from the logarithm. Work backwards.

1. Let the answer be u + iv.
2. Apply the forward function to both sides, so the inverse disappears.
3. Separate the forward function of u + iv into real and imaginary parts, using the results already proved.
4. Compare parts with the given number. You now have two real equations in u and v.
5. Solve them, usually by taking a ratio or by using an identity.

The whole trick is step two: an inverse function is hard to manipulate and its forward partner is easy, so get rid of the inverse first.

Worked: tanh inverse of x + iy

Let u + iv be the answer, so that x + iy is the hyperbolic tangent of u + iv.

Now use the separation of tanh from the chapter on the tangent of a complex angle.

Comparing parts gives the two equations below.

Now the standard manoeuvre. Form x squared plus y squared, and separately form 2x over (1 minus x squared minus y squared). The algebra is routine and the results are these.

So:

Those two are the answer, and they are the pair MU's own reading list prints. Notice the symmetry: the same expression appears in both with x and y exchanged and one sign changed, which is a good way to recall them.

Worked: sin inverse of a complex number

Let u + iv be sin inverse of (a + ib), so a + ib is the sine of u + iv.

Use the separation of the sine.

Comparing parts:

Two equations, two unknowns. Eliminate v by using cosh squared minus sinh squared equals one: divide the first by sin u and the second by cos u, then square and subtract.

In one line

Zero over zero is not a number and not an error: it is a signal that the limit has to be worked out another way.

Why this chapter is in the book

MU's sixth Course Objective for this paper reads: "Inculcate the habit of Mathematical Thinking through Indeterminate forms". No module label mentions indeterminate forms at all.

That is recorded in this book's own contract rather than glossed over, and the topic is taught for two reasons. The internal twenty marks includes five for quizzes and assignments, which are set against the Course Objectives. And the topic is not decoration in this paper: the moment you have to show that the defining integral of a Laplace transform converges, or evaluate an initial-value theorem, or handle a partial fraction with a repeated root, you are taking a limit that arrives as an indeterminate form.

What "indeterminate" means

A form is indeterminate when knowing only the limits of the pieces does not tell you the limit of the whole.

Take a fraction whose top tends to zero and whose bottom tends to zero. What does the fraction do? The honest answer is: anything at all, depending on which of them gets to zero faster. Here are three fractions, all of the form zero over zero, with three different answers.

Four fractions of the same form, and the answers are 1, 2, 0 and unbounded. So the form tells you nothing, which is exactly what indeterminate means.

Compare that with a form that is not indeterminate. If the top tends to 5 and the bottom tends to 0 from above, the fraction grows without limit, and there is nothing to work out. That form is determinate.

The seven indeterminate forms

Those seven are the whole list. Everything else is determinate.

In one line

For a limit of the form zero over zero or infinity over infinity, differentiate the top and the bottom separately and try again.

The rule

Suppose f and g both tend to zero as x tends to a, or both grow without limit. Then the limit of f over g equals the limit of f prime over g prime, provided that second limit exists.

That proviso matters, and so do two others.

Differentiate them separately. This is not the quotient rule. The top is differentiated, the bottom is differentiated, and the two results are put over each other. Using the quotient rule here is the commonest mistake with L'Hopital, and it gives nonsense.

Check the form first. The rule applies only to zero over zero and infinity over infinity. Applying it to a determinate form gives a wrong answer with great confidence, and the next section shows that happening.

You may repeat it. If the new fraction is still indeterminate, differentiate again. Some limits need three rounds.

Worked

Take the second. At x = 0 the top is 1 minus 1, which is zero, and the bottom is zero, so the form is zero over zero. Differentiate: the top becomes sin x and the bottom becomes 2x, and the form is still zero over zero. Differentiate again: the top becomes cos x and the bottom becomes 2, and now putting x = 0 gives one over two.

The fourth needs three rounds and ends with cos x over 6 evaluated at zero, giving one sixth.

And the infinity over infinity cases.

The third takes three rounds: each differentiation drops the power of x by one and leaves the exponential alone, so after three the top is a constant and the bottom still grows.

That is the proof of a fact stated in the previous chapter: an exponential beats any power of x. It is the reason the Laplace integral converges.

Applying it where it does not apply

Here is the failure, in the smallest case, so that it is met once on paper rather than in an examination.

Consider the limit of (x + 3) over (2x + 1) as x tends to 1. The top tends to 4 and the bottom tends to 3, so the form is 4 over 3 and the limit is 4 over 3. There is nothing to work out.

In one line

A transform carries a hard problem into a place where it is easy, you solve it there, and you carry the answer back.

MU's label

Her label is one word: "Introduction". This chapter is what belongs under it, and it comes before any integral, because the integral makes no sense until you know what it is for.

The pattern, with an example you already know

Suppose you had to multiply 4,096 by 8,192 by hand, and suppose you had a book of logarithms and no calculator. You would not multiply. You would do this.

Multiplication, which is hard by hand, became addition, which is easy. The logarithm carried the problem into a place where the operation was simpler, and the antilogarithm carried the answer back.

That is the whole idea of a transform, and every transform in mathematics follows that three-step shape.

The Laplace transform's version of it

The hard operation here is differentiation, and the problem is a differential equation.

A differential equation goes in, an ordinary algebraic equation comes out, you solve it with school algebra, and you convert back. That is what the whole second half of Module 1 is for.

The step that makes it work is one theorem, the transform of a derivative, which has its own chapter. It says that transforming dy/dt gives s times the transform of y, minus the value of y at the start. Differentiation becomes multiplication, and the initial condition walks into the algebra by itself.

Why that is worth learning when Module 2 solves the same equations by hand

A fair question, and there are three honest answers.

The initial conditions come for free. By the methods of Module 2 you find a general solution with arbitrary constants and then substitute the initial values to pin them down, which is a second stage of work and a second chance to make a mistake. The transform method puts the initial values in at the start and never produces an arbitrary constant at all.

A discontinuous input is no harder. If the right-hand side of the equation switches on at t = 2, or is a sudden impulse, the methods of Module 2 have to solve the equation in pieces and match the pieces at the joins. The transform method handles it with one extra factor, and the chapter on equations driven by a step or an impulse shows it.

In one line

The Laplace transform of f(t) is the integral from zero to infinity of e to the minus st times f(t) dt, and it is a function of s.

The definition

Figure 40.1 MU's own label is the definition of the Laplace transform, and her Outcome 2 says a student must be able to find a transform 'using definition'. So this integral is the centre of the next twenty chapters, not a formality to be skipped on the way to the table.

How an integral is written in the running text of this book

The integral sign cannot be typeset in a body here, so an integral is written the way it is typed at a keyboard, and it is worth reading the convention once because it is used in every chapter that follows.

integrate(f, x) means the integral of f with respect to x.

integrate(f, (t, 0, oo)) means the definite integral of f with respect to t, from 0 to infinity. The bracket holds the variable first and then the two limits, lower before upper, and oo is infinity.

So the defining integral in the figure above, written that way, is integrate(e^(-s t) f(t), (t, 0, oo)). Every such line in this book has been evaluated by machine and checked against the answer printed beside it.

Four things to read off the definition, and all four are examined.

It is a definite integral, so the answer contains no t. The variable t is integrated away between the limits, exactly as the x disappears when you evaluate a definite integral in x.

The answer is a function of s. That is the whole point: a function of time has been turned into a function of a different variable. The notation is f(t) for the original, always lower case, and F(s) for its transform, always capital. That convention is used everywhere and you should adopt it.

The lower limit is zero, not minus infinity. So the transform knows nothing at all about what f did before time zero. This is deliberate and it is why the transform suits problems that start when you switch something on.

The kernel is e to the minus st. That factor is the entire mechanism. It decays as t grows, and s controls how fast, so the integral only ever sees the part of f near the start unless s is small.

The notation

MU writes the transform operator as L. Some books write a script capital L, and some write it with square brackets. All mean the same.

In one line

The transform exists if f is piecewise continuous and does not grow faster than an exponential, and then only for s large enough.

The question

The definition is an integral to infinity. Such an integral may fail to exist, and the two conditions below are what guarantee it does not. A question asking for "the sufficient conditions for the existence of the Laplace transform" wants them stated, so they are worth learning as sentences.

Condition one: piecewise continuity

f must be piecewise continuous on every finite interval from 0 onwards. That means: on any finite stretch it has at most finitely many breaks, and at each break the function jumps by a finite amount rather than shooting off.

A square wave is piecewise continuous. The unit step function is piecewise continuous. The function one over t is not, because at t = 0 it does not jump by a finite amount, it grows without limit.

The reason the condition is needed: a finite number of finite jumps can be integrated over by splitting the integral at the jumps, and a finite jump contributes nothing to the area. An infinite blow-up cannot be handled that way.

Condition two: exponential order

f must be of exponential order. That means there are numbers M and a such that the size of f(t) is at most M times e to the at for all large t.

In words: f may grow, but not faster than some exponential. Every function in this paper satisfies this, and it is worth seeing which functions do and which do not.

The claim in the second row is the one from the indeterminate-forms chapter, and it is why every polynomial has a transform.

Why the two conditions do the job

Because together they make the integrand small enough, for large enough s.

The size of the integrand is at most e to the minus st times M e to the at, which is M times e to the minus (s minus a)t. The integral of that from zero to infinity converges whenever s minus a is positive.

So the transform exists for s greater than a, and that is where the condition on s in every table entry comes from. It is not a technicality bolted on afterwards; it is the number a from the exponential-order condition.

In one line

The transform of t to the n is n factorial divided by s to the n plus one, and it comes out of integrating by parts n times.

The transform of 1

Done in the definition chapter, and repeated here because this chapter is the table's first three rows.

The working: the antiderivative of e to the minus st in t is minus e to the minus st over s. At infinity that tends to zero when s is positive; at zero it is minus one over s. Subtracting gives one over s.

The transform of t

Integrate by parts, taking t as the part to differentiate and e to the minus st as the part to integrate.

The working, which is what a question wants. By parts, the integral equals the boundary term, t times minus e to the minus st over s, evaluated from zero to infinity, plus one over s times the integral of e to the minus st.

The boundary term is zero at both ends: at t = 0 because of the factor t, and at infinity because the exponential beats the t, which is the limit the indeterminate-forms chapter proved.

So the whole thing is one over s times the transform of 1, which is one over s squared.

The transform of t squared, and the pattern

The same integration by parts, now with t squared, leaves two over s times the transform of t.

The pattern is visible. Each step down brings out a factor and raises the power of s by one.

The general result, by induction

Integrating by parts once, with t to the n, gives the relation below between consecutive transforms.

The boundary term vanishes at both ends for the same two reasons as before. Applying that relation n times, and finishing with the transform of 1, gives the following.

That is the general result, and it is the row of the table you will use most. Valid for s greater than zero, and for n a non-negative whole number.

Note that n = 0 gives the transform of 1, since zero factorial is one, so the general formula contains the first case.

A non-whole power

If n is not a whole number, the factorial has no meaning, and it is replaced by the gamma function, which extends the factorial to non-whole arguments.

In one line

The transform of e to the at is one over s minus a, valid for s greater than a, and the condition is the whole content of the result.

The derivation

One line of integration, and it is worth doing in full because every later shifting theorem is this calculation in disguise.

The integrand is e to the minus st times e to the at, and multiplying exponentials adds the indices.

So the integral is exactly the one that gave the transform of 1, with s replaced by s minus a.

Therefore:

The antiderivative is minus e to the minus (s minus a)t over (s minus a). At infinity it tends to zero only if s minus a is positive; at zero it is minus one over (s minus a). Subtracting gives the result.

The condition, which is the point

The transform is valid for s greater than a and for nothing less.

That is not a footnote. It says: the faster f grows, the further to the right you must be for the integral to exist. If f grows at rate 3, the kernel must decay faster than rate 3, which needs s above 3.

And the boundary shows in the answer itself: the denominator s minus a vanishes exactly at s = a, which is the edge of the region of convergence. A transform's denominator tells you where it stops being valid, which is a useful reading habit.

The two signs

This is where marks are lost, so it is worth labouring.

A positive exponent gives a minus in the denominator. A negative exponent gives a plus. The rule that never fails: the denominator is s minus (the number in the exponent), including its sign.

The last row is the transform of 1 again, so this result contains the first one.

Complex a, which is used immediately

Nothing in the derivation required a to be real. If a is complex the result still holds, provided the real part of s exceeds the real part of a.

That is worth having, because the next chapter computes the transform of a sine by writing it as a combination of two complex exponentials and using this result twice. Module 1's first half pays for its second half there.

In one line

The transform of sin at is a over s squared plus a squared, and of cos at is s over s squared plus a squared, and the quickest derivation goes through complex exponentials.

The two results

Both valid for s greater than zero, since a sine and a cosine are bounded, so their exponential order is zero.

Remembering which is which: the sine has a constant on top, the cosine has an s on top. And the denominators are identical. A useful check: at s = 0 the cosine transform is zero, which is right because integrating a cosine over all time gives nothing net.

The derivation with complex exponentials

This is the short route, and it is the first place the two halves of Module 1 meet.

The Euler chapter gave the sine and cosine as combinations of exponentials.

The previous chapter gave the transform of an exponential, for complex exponents as well as real ones. Apply it to each piece.

Now the cosine is half the sum of those two.

Adding the two fractions gives 2s over (s minus ia)(s plus ia), and the denominator is s squared plus a squared because the i squared turns minus a squared into plus a squared. Halving gives the result.

The sine is the difference divided by 2i.

The difference is 2ia over (s squared plus a squared), and dividing by 2i leaves a over the same denominator.

The derivation by parts, for a question that insists

An examination sometimes asks for the transform "from the definition", meaning without complex numbers. Integrate by parts twice and an equation for the transform appears.

Integrating e to the minus st times sin at by parts twice returns the same integral multiplied by minus a squared over s squared, plus a boundary contribution of a over s squared. Calling the transform F, that says F equals a over s squared minus a squared F over s squared, so F times (1 plus a squared over s squared) is a over s squared, and hence F is a over (s squared plus a squared).

That is the whole of the working, and it is worth writing out once by hand. The trick, "integrate by parts twice and solve for the integral", is used again in the periodic-function chapter.

In one line

Same shape as the sine and cosine, with a minus sign in the denominator instead of a plus.

The two results

Valid for s greater than the size of a, because sinh and cosh grow at rate a.

The derivation, which is two lines

The hyperbolic functions are combinations of real exponentials, so the transform of an exponential is all that is needed.

Transform each piece and combine.

The difference of the two fractions is 2a over (s minus a)(s plus a), and the sum is 2s over the same, and that denominator is s squared minus a squared. Halving gives the two results.

The sign, which is the whole difference

Set it out side by side, because this is what is examined.

The reason for the difference, in one sentence: the circular pair are combinations of imaginary exponentials, so squaring the i turns minus into plus, and the hyperbolic pair are combinations of real exponentials, so nothing changes sign.

You can also get one pair from the other with the relations of Module 1's first half. The transform of sinh at is the transform of sin(iat) divided by i, and replacing a by ia in the sine's answer turns a squared into minus a squared.

Why the region of convergence is different

For the circular pair, s greater than zero. For the hyperbolic pair, s greater than the size of a.

That difference is real and it matters: a hyperbolic function grows, so the kernel has to decay faster to beat it. And it shows in the answer, because s squared minus a squared vanishes at s = a, which is the boundary.

Inverting, which is where the minus sign earns its keep

A denominator that factorises over the reals means a hyperbolic answer or a pair of exponentials; a denominator that does not means a circular answer.

Two of those denominators look almost identical and give completely different functions, one growing and one oscillating. Reading the sign of the constant term is therefore the first thing to do with any quadratic denominator.

A useful equivalence: the hyperbolic answers can always be written as two exponentials instead, since the denominator factorises.

In one line

These fourteen rows, with their conditions on s, are what the rest of the module reads in both directions.

MU's label

"Table of Elementary Laplace Transforms". Every row below was derived from the definition in one of the previous six chapters, not copied from anywhere, and every one was recomputed by machine before it was printed.

The table

The first eight rows are the ones derived so far. Rows nine to twelve come from the first shifting theorem, and thirteen and fourteen from the rule for multiplying by t, both of which are the next few chapters. They are printed here so that the table is in one place.

Every row, verified:

How to read it forwards

Match the shape of f, read off the F, and substitute the numbers. Use linearity to break a sum into its terms.

How to read it backwards, which is harder

Going from F to f needs the F to be made to look like a row of the table, and that is where the work is. Three manoeuvres do almost all of it.

Supply a missing constant. The sine row has an a on top. If the numerator is 1, write it as one over a times a.

In one line

The transform of a sum is the sum of the transforms, and a constant passes straight through, but nothing else is respected.

The property

This is the first of MU's "Theorems on Important Properties of Laplace Transformation", and the one used in every single question. Her label covers seven results in all, and the table at the end of the chapter on multiplying by t gathers the properties of the Laplace transform in one place, each with its own chapter: linearity, the two shifting theorems, multiplication by a power of t, division by t, change of scale, the transform of a derivative and the transform of an integral.

In words: the Laplace transform is a linear operator.

Why it holds

Because integration is linear, and the transform is an integral. The integral of a sum is the sum of the integrals, and a constant factor comes out in front. There is nothing more to the proof than that, and an examination answer can be three lines long.

Write the transform of the combination as its defining integral, split the integral at the plus sign, and take the constants out. Each remaining integral is one of the two transforms by definition.

Worked

The last line needs the bracket expanded before linearity can be used: 4 plus 4t plus t squared. Linearity handles sums, and a square is not a sum until you make it one.

What linearity does NOT give you

This is the real content of the chapter, and it is the source of more wrong answers than any other single thing in the module.

None of those is true. Here is the counterexample, on the simplest possible functions. Take f(t) = g(t) = t.

The product of the two transforms would be one over s to the fourth. The transform of the product is 2 over s cubed. They are not equal and they are not even the same shape.

So whenever a product appears, one of three things has to happen.

Turn the product into a sum first, with an algebraic or trigonometric identity. This is what to try first, and it works surprisingly often.

Use a shifting theorem, if one factor is an exponential. The next chapter.

In one line

Multiplying f(t) by e to the at shifts the transform: replace s by s minus a everywhere.

The theorem

MU's label is "First Shifting Theorem".

In words: if you know the transform of f, then the transform of e to the at times f is the same function of s with s replaced by s minus a.

It is also called the shifting property in s, or the first translation theorem.

The proof, which is one line

Write the transform of e to the at f(t) as its integral. The two exponentials combine.

So the integral is the defining integral of the transform of f, with s minus a in place of s. That is F(s minus a), and the proof is finished.

The condition on s shifts with it: if F(s) was valid for s greater than c, then F(s minus a) is valid for s greater than c plus a.

What it buys: eight new table rows, with no integration

Every row of the elementary table can be multiplied by an exponential, for free.

Every one of those, verified:

The one place people go wrong

The substitution is everywhere in F, not just in one place.

Take the transform of cos 4t, which is s over (s squared plus 16). Multiplying by e to the 3t means replacing s by s minus 3 in both the numerator and the denominator.

Replacing it only in the denominator, and leaving an s on top, is the standard error.

The two are different functions, and the checker proved the equality false, which is exactly what a student's own check should do: put s = 3 into both. The left side is zero; the right side is three sixteenths.

In one line

Every transform of the form e to the at times something is this theorem, and every inverse of a completed square is it read backwards.

Forward, a graded set

Each of these is the same two steps: transform the function, then replace s by s minus a.

The fifth and sixth show the order that matters: linearity first, inside the bracket, and the shift applied to the whole result.

A product of two exponentials, which is not a special case

Combine the exponentials before doing anything else. Applying the shifting theorem twice would also work and gives the same answer, but combining is quicker and less error-prone.

A product of an exponential and a hyperbolic function

Alternatively, write the hyperbolic function as two exponentials and combine each with the first, which avoids the theorem altogether.

Both answers are correct and they are equal. Which form you leave it in depends on what comes next: the second is already in partial fractions, which is convenient if it is going to be inverted.

Backwards: the recognition that makes it work

The theorem read in reverse says: if every s in F(s) appears as (s minus a), pull out an e to the at.

The three shapes to recognise:

with n a positive whole number; and, with the quadratic completed,

Those three rows cover nearly every inverse in this paper that is not a plain partial fraction.

Worked inverse, with the completing of the square shown

Invert the function below.

The denominator does not factorise over the reals, because its discriminant, 36 minus 52, is negative. So complete the square.

Now every s appears as s plus 3, so a is minus 3 and b is 2.

The one over two comes from the sine row needing a b on top, and b is 2.

In one line

Multiplying f(t) by t differentiates the transform and changes its sign.

The theorem

In words: the transform of t times f is minus the derivative, with respect to s, of the transform of f.

This is one of MU's "Theorems on Important Properties", and it is the companion of the first shifting theorem: that one handles multiplication by an exponential, this one handles multiplication by t.

Why the minus sign is there

Differentiate the defining integral with respect to s. The only s in the integrand is in the kernel e to the minus st, and differentiating that with respect to s gives minus t times e to the minus st.

So differentiating F(s) produces an extra factor of minus t inside the integral, which is the transform of minus t f(t). Multiplying through by minus one gives the theorem.

That is the whole proof, and it is worth knowing because it explains the sign rather than asking you to remember it. The minus sign comes from the minus in the exponent of the kernel.

Worked, the entries it gives

The transform of e to the at is one over (s minus a). Differentiating with respect to s gives minus one over (s minus a) squared, and negating gives the answer. Notice that the first shifting theorem gives the same row, which is a useful cross-check: two independent routes, one answer.

For the first: the transform of sin at is a over (s squared plus a squared). Differentiating with respect to s gives minus 2as over (s squared plus a squared) squared, and negating gives the printed answer.

For the second: differentiating s over (s squared plus a squared) needs the quotient rule and gives (a squared minus s squared) over the square of the denominator, and negating flips it to s squared minus a squared.

These two rows are the ones the shifting theorem cannot reach, so they have to be got this way, and they are set regularly.

Worked, a longer one

Find the transform of t times e to the 2t times sin 3t.

Two ways, and both are instructive.

Route one: shift, then differentiate. The transform of e to the 2t sin 3t is 3 over ((s minus 2) squared plus 9). Differentiate with respect to s and negate.

In one line

Multiplying by t to the n differentiates the transform n times, and the sign alternates.

The theorem

Differentiate F n times with respect to s, and multiply by minus one to the n, which is plus for even n and minus for odd n.

Why, by induction

The previous chapter proved the case n = 1. Suppose the result holds for n. Then t to the n plus one times f is t times (t to the n times f), so applying the n = 1 rule to it differentiates once more and brings in one more minus sign. That is the step, and the base case is n = 0, where the statement says F equals F.

Worked

For the first: the transform of e to the at is one over (s minus a). Differentiating twice gives 2 over (s minus a) cubed, and minus one squared is plus one, so the sign is unchanged.

The general row is the one worth carrying, and note that it is the same row the first shifting theorem gives when applied to the transform of t to the n. Two routes, one answer, which is again a useful cross-check.

The harder ones, which is what this theorem is for

The shifting theorem cannot produce these, so they must come from here.

Each is two differentiations of a quotient, which is tedious but mechanical. The way to keep it under control is to differentiate once, simplify fully, and only then differentiate again; differentiating twice without simplifying in between produces an expression nobody can check.

Worked, for the sine with a = 1, so the algebra is visible.

Then minus one squared is plus one, so the transform of t squared sin t is (6 s squared minus 2) over (s squared plus 1) cubed.

The companion rule, which has its own chapter

MU's properties list also reaches the rule for f(t) divided by t, and it is the mirror image of this one: where multiplying by t differentiates the transform, dividing by t integrates it. That rule, its proof and the standard transforms it produces are the next chapter, because its worked examples are of a different kind: they produce logarithms and arctangents rather than rational functions.

In one line

The transform of f(t) over t is the integral of F from s to infinity.

The theorem

The variable of integration is written u to keep it distinct from the s that remains as the answer's variable.

The condition: the limit of f(t) over t as t tends to zero must exist. That is what stops the integral near zero from misbehaving, and it is the condition to state in an answer.

The proof

Start from the definition of F and integrate both sides with respect to s, from s to infinity.

The right-hand side becomes a double integral, in u and in t. Reverse the order: do the u integral first, with t held fixed.

That is where the one over t comes from: integrating the kernel with respect to its own parameter divides by t. What is left is the defining integral of the transform of f(t) over t, which proves the theorem.

Reversing the order of a double integral is the same move that proves the convolution theorem later, and it is worth being comfortable with.

Worked: the transform of sin t over t

This is the standard question on this rule.

The transform of sin t is one over (u squared plus 1). Integrate it with respect to u from s to infinity.

The antiderivative is the arctangent of u. At infinity it is pi over two; at u = s it is the arctangent of s.

So the answer is the arctangent of one over s.

The condition is satisfied: sin t over t tends to 1 as t tends to zero, which is the limit the indeterminate-forms chapter computed.

And a bonus. Putting s = 0 in the answer gives pi over two, and putting s = 0 in the defining integral gives the integral of sin t over t from zero to infinity. So this transform computes a famous integral as a by-product.

Worked: the transform of (1 minus cos at) over t

The transform of 1 minus cos at is one over u minus u over (u squared plus a squared). Integrating with respect to u gives log u minus half the log of (u squared plus a squared), which is the log of u over the square root of (u squared plus a squared).

At infinity that logarithm tends to log 1, which is zero, because u and the square root behave the same way for large u. At u = s it is the log of s over the square root of (s squared plus a squared). Subtracting gives minus that, which is the log of the square root over s, and that is the printed answer.

In one line

Stretching time by a factor a divides the transform by a and divides s by a too.

The theorem

for a positive. Both changes happen together, and dropping either one is the mistake.

The proof

In the defining integral, substitute u = at, so t is u over a and dt is du over a. The kernel becomes e to the minus (s over a)u, and the extra one over a comes out in front.

That is the whole proof and it is one substitution.

Worked

Check the theorem on those two. The transform of sin t is one over (s squared plus 1). Replacing s by s over 3 gives one over (s squared over 9 plus 1), which is 9 over (s squared plus 9). Dividing by 3 gives 3 over (s squared plus 9), which is the first line.

That is the theorem verified on a case where the answer is independently known, which is the right way to meet it.

What it is for

Honestly: not very much, in this paper. Every transform you need can be got from the elementary table plus the shifting theorems. The change-of-scale rule is on MU's list of properties, so it is examinable, and it occasionally saves work on a transform whose stretched form you happen to know.

Where it does earn its place is with a transform that is not a table row.

Getting the second from the first by the rule takes one line, where repeating the integration of the previous chapter takes several. Note that the function has to be written as sin(2t) over (2t), so that it really is f(2t) with f being sin t over t.

And multiplying by 2 gives the more usual form.

The mirror rule on the transform side

There is a companion statement, sometimes asked, about scaling s instead of t.

Same shape, read the other way, and proved by the same substitution.

That inverse can be got either by the mirror rule or, more simply, by dividing top and bottom by 2 and using the exponential row, which is what most students do and which is perfectly correct.

The trap

The rule needs a positive. For a negative it is false, because the substitution reverses the limits of the integral, and in any case f(at) for negative a would look at f at negative times, about which the transform knows nothing.

So a question about f(minus t) has no answer within this theory. If it appears, it is a question about the lower limit of the defining integral being zero.

In one line

The transform of f prime is s times F(s) minus f(0), which is what turns a differential equation into an algebraic one.

The theorem

MU's label is "Laplace Transform of Derivatives", and this is the most important single result in the paper.

Two things have happened. Differentiation has become multiplication by s. And the initial value has walked into the answer by itself.

The proof

Integrate the defining integral by parts, taking f prime as the part to integrate and the kernel as the part to differentiate.

The boundary term is e to the minus st times f(t), evaluated from zero to infinity. At infinity it is zero, because f is of exponential order and the kernel beats it for large enough s. At zero it is f(0).

So the boundary term contributes minus f(0). The remaining integral is s times the defining integral of F.

That minus s, moved to the other side, is where the plus s F(s) comes from. The proof is four lines and is worth being able to write.

The second derivative

Apply the theorem to f prime instead of f.

More carefully, with the initial values named: the transform of the second derivative is s squared F(s) minus s times f(0) minus f prime of zero.

The pattern is now visible. Each differentiation brings another factor of s, and each brings in one more initial value, with the powers of s counting down.

The nth derivative

The transform of the nth derivative is s to the n times F(s), minus s to the n minus one times f(0), minus s to the n minus two times f prime of zero, and so on, ending with minus the (n minus one)th derivative at zero.

For this paper you need the first two, and occasionally the third.

Note how many initial values each needs: one for the first derivative, two for the second, three for the third. That matches what Module 2 says about a differential equation of order n needing n conditions to pin down a particular solution, and it is the same fact seen from the transform side.

Why this makes the whole method work

Take a differential equation. Every term is either a derivative of y, or y itself, or a known function of t. Transform every term.

• y becomes Y(s).
• y prime becomes sY minus y(0).
• y double prime becomes s squared Y minus s y(0) minus y prime of zero.
• the known function becomes its own transform.

In one line

The transform of the integral of f from 0 to t is F(s) divided by s.

The theorem

Where differentiation multiplied by s, integration divides by s. That symmetry is the cleanest single thing about the Laplace transform.

The proof, in one line from the previous chapter

Let g(t) be the integral of f from 0 to t. Then g prime is f, by the fundamental theorem of calculus, and g(0) is zero, because the integral from 0 to 0 is zero.

Apply the derivative theorem to g.

So the transform of f equals s times the transform of g minus zero, and dividing by s gives the result.

That is the whole proof, and it shows how much work the derivative theorem does: this second theorem is a corollary of it.

Worked

The integral of 1 from 0 to t is t, so the theorem says the transform of t should be the transform of 1 divided by s, which is one over s squared. It is.

The integral of t from 0 to t is t squared over 2, so the transform should be one over s squared, divided by s, which is one over s cubed. It is.

The integral of sin u from 0 to t is 1 minus cos t, so the theorem says its transform is one over s(s squared plus 1). Check that against the first line.

They agree.

Reading it backwards, which is the common use

The reverse reading is: a factor of one over s in F(s) means an integration in t.

Each of those could also be done by partial fractions, and usually is. The integration route is quicker when the rest of the fraction is a clean table row and you would rather integrate once than split into three fractions.

Worked, the first one. Strip off the one over s. What is left is one over (s squared plus 4), whose inverse is sin 2t over 2. Integrate that from 0 to t.

The integral of sin 2u over 2 is minus cos 2u over 4, evaluated from 0 to t, which is (1 minus cos 2t) over 4. That is the answer.

Repeated integration

Two factors of one over s means integrating twice.

Strip off one over s squared. What is left is one over (s squared plus 1), whose inverse is sin t. Integrating sin once gives 1 minus cos t, and integrating that gives t minus sin t.

In one line

The value of f at the very start is the limit of s F(s) as s grows, and the value it settles down to is the limit of s F(s) as s tends to zero.

The two theorems

The first is the initial value theorem and the second the final value theorem. Both are on MU's list of important properties.

What they are for

They let you read off the beginning and the end of the answer without inverting the transform at all. In a problem where you only want to know where a system starts and where it settles, that saves the whole inversion.

They are also a check on an inversion you have already done: if the answer's value at zero does not match what the initial value theorem says, one of the two is wrong.

The initial value theorem, and where it comes from

From the derivative theorem: the transform of f prime is sF(s) minus f(0). As s grows, the transform of any function tends to zero, because the kernel decays faster and faster. So sF(s) minus f(0) tends to zero, which is the theorem.

Worked.

Read those off. The inverse of one over (s plus 3) is e to the minus 3t, whose value at zero is 1: the theorem agrees. The inverse of one over s squared is t, whose value at zero is 0: agrees. The inverse of s over (s squared plus 4) is cos 2t, whose value at zero is 1: agrees.

There is a quick way to see the answer for a rational F: compare the degrees. If the numerator's degree is exactly one less than the denominator's, sF(s) tends to the ratio of the leading coefficients. If the gap is two or more, the limit is zero.

The final value theorem, and the condition it needs

The first says e to the minus 3t settles to 0, which it does. The second says the inverse of one over s(s plus 2), which is (1 minus e to the minus 2t) over 2, settles to one half, which it does.

The condition matters and the theorem is false without it. The final value theorem holds only when f actually settles down to a limit. For a rational F that means every root of the denominator has a negative real part, with at most a single root at zero.

In one line

The unit step is zero before a and one after it, and it is how you write "switches on at time a" in a formula.

Where this chapter sits, and why it is not where MU puts it

MU files the Heaviside unit step function under her 1.3, among the inverse-transform topics. Nothing about it is an inverse transform, and the second shifting theorem, which she puts in 1.2, cannot even be stated without it. So this book teaches it here, in the forward-transform section, and the plan records the move. It stays inside Module 1 either way.

The definition

The unit step function, or Heaviside function, written u(t minus a), is zero for t less than a and one for t greater than a.

Figure 57.1 The unit step on the left. The right-hand picture is the pulse of unit area used to build the impulse function two chapters from here; both are drawn together because the impulse is the derivative of the step.

The value at t = a is a matter of convention and never matters here, because a single point contributes no area to an integral. Some books set it to one half.

The simplest case, a = 0, is written u(t) and is 1 for all positive t. So the constant function 1 and the unit step u(t) are the same thing as far as the Laplace transform is concerned, since the transform never looks at negative t.

Its transform

The general result:

The derivation is one line. The integrand is zero up to a, so the integral runs from a to infinity instead of from zero. Substituting v = t minus a turns it into the transform of 1 multiplied by e to the minus as.

Note what appears: an exponential in s. That is the signature of a shift in time, and the chapter on the second shifting theorem makes it general. Any time you see an e to the minus as in a transform, something is being switched on at time a.

What it is for: switching

A function multiplied by u(t minus a) is that function with everything before time a erased.

In one line

Delaying f by a multiplies its transform by e to the minus as, provided the delayed function is switched on with a unit step.

The theorem

MU's label is "Second Shifting Theorem".

Compare the two shifting theorems, because the pairing is the point.

The first shifts in s; the second shifts in t. That is why they are named in that order.

The proof

Write the defining integral. The unit step makes the integrand zero up to t = a, so the integral runs from a to infinity.

Substitute v = t minus a, so t is v plus a and dt is dv. The kernel becomes e to the minus s(v plus a), which splits into e to the minus as times e to the minus sv, and the e to the minus as is a constant that comes out in front.

What is left is the defining integral of F in the variable v. So the answer is e to the minus as times F(s), and the proof is three lines.

Why the unit step has to be there

This is the part students skip and then get wrong, so it is worth being blunt.

The theorem as stated is about the function that is zero before a and equals f(t minus a) after it. Without the step factor, f(t minus a) would be some non-zero thing for t less than a, and the integral from 0 to a would contribute something the theorem does not account for.

So "delayed by a" always means "delayed by a and zero before that". If your function is not zero before a, the theorem does not apply and you must write the function out as a sum of pieces first.

Worked

Here f is the constant 1, whose transform is one over s, delayed by 2. The theorem gives e to the minus 2s over s, which is the result the previous chapter derived directly.

In each case: identify f from the shape, look up F, multiply by e to the minus as. The pattern to spot is that the same a appears inside the function and inside the step, which is what tells you the theorem applies directly.

In one line

At each break point, add a step multiplied by the change in the formula there.

Why this chapter exists

The second shifting theorem is useless until the function is written as a single expression in unit steps. Students learn the theorem, meet a function defined in three pieces, and stop. This is the missing step, and it is purely mechanical.

The rule

Suppose f is given by formula g1 on the first interval, g2 on the second, g3 on the third, with break points at a and b. Then:

f equals g1, plus (g2 minus g1) times u(t minus a), plus (g3 minus g2) times u(t minus b).

In words: start with the first formula, and at each break add a step multiplied by the jump from the old formula to the new one.

The reason it works: before a, every step is zero, so only g1 survives. Between a and b, the first step is one, so you have g1 plus g2 minus g1, which is g2. After b, both steps are one, so you have g3. Each correction switches on exactly where it is needed and cancels exactly what it should.

Worked: a function that switches on

Let f be 0 for t less than 3, and 5 for t greater than 3.

The first formula is 0. The jump at 3 is 5 minus 0, which is 5.

So f equals 5u(t minus 3).

Worked: a function that steps up

Let f be 2 for t less than 1, and 7 for t greater than 1.

First formula 2, jump at 1 is 5.

So f equals 2 plus 5u(t minus 1).

Worked: a pulse

Let f be 0 before 1, then 4 between 1 and 3, then 0 again after 3.

First formula 0. Jump at 1 is plus 4. Jump at 3 is 0 minus 4, which is minus 4.

So f equals 4u(t minus 1) minus 4u(t minus 3).

That is the window pattern, and every pulse is a difference of two steps.

Worked: a ramp that stops

Let f be t for t less than 2, and 2 for t greater than 2. So it rises with slope one and then holds.

First formula t. The new formula is 2, so the jump at t = 2 is 2 minus t, which is minus (t minus 2).

So f equals t minus (t minus 2)u(t minus 2). Note how neatly the jump came out already in the (t minus a) form the second shifting theorem wants.

In one line

Integrate over one period only, then divide by one minus e to the minus pT.

The theorem

If f has period T, meaning f(t plus T) equals f(t) for every t, then:

So you never integrate over the whole infinite range. One period is enough, and the denominator accounts for all the rest.

The proof, which is a geometric series

Split the defining integral at every multiple of T: from 0 to T, from T to 2T, from 2T to 3T, and so on.

In the nth piece, substitute t = v plus nT. Because f is periodic, f(v plus nT) is just f(v). The kernel becomes e to the minus s(v plus nT), which is e to the minus snT times e to the minus sv.

So every piece is the same integral, the one over the first period, multiplied by e to the minus snT. Adding them up gives that integral multiplied by the sum of a geometric series with ratio e to the minus sT.

The series 1 plus r plus r squared and so on sums to one over (1 minus r) when r is less than one in size, and e to the minus sT is less than one for positive s. That gives the theorem.

Worked: the square wave

Let f be 1 for the first half of each period and minus 1 for the second half, with period 2a.

The integral over one period splits into two. From 0 to a the integrand is e to the minus st; from a to 2a it is minus e to the minus st.

Doing both and combining gives (1 minus e to the minus as) squared over s, and dividing by 1 minus e to the minus 2as, which factorises, leaves the following.

With a = 1 for clarity that is (1 minus e to the minus s) over s(1 plus e to the minus s), and that can be tidied with the definition of the hyperbolic tangent.

So the transform of the square wave of period 2 is the hyperbolic tangent of s over 2, divided by s. A hyperbolic function appearing from a square wave is the kind of thing worth remembering, and it is a standard examination answer.

Worked: the sawtooth

Let f(t) equal t on the interval from 0 to a, repeating with period a.

The integral over one period is the integral of t e to the minus st from 0 to a, which is done by parts.

In one line

The unit impulse is the limit of a tall narrow pulse of area one, its transform is e to the minus as, and it is the mathematics of a hammer blow.

Building it

Take a rectangular pulse that starts at t = a, has width h, and has height one over h, so its area is exactly one whatever h is. The right-hand picture in the unit step chapter is that pulse.

Now let h shrink. The pulse becomes narrower and taller, and its area stays one. In the limit it is zero everywhere except at the single point a, where it is infinitely tall, and it still has area one.

That object is the Dirac delta function, or unit impulse function, written as delta(t minus a). MU's label is "Dirac-delta Function (Unit Impulse Function)".

What kind of object it is, said honestly

It is not a function. No function can be zero everywhere except at one point and still enclose area one, because a single point has no width. The delta is a distribution, or generalised function, defined by what it does inside an integral rather than by its values.

What it does is this: integrating any continuous g against delta(t minus a) picks out the value of g at a.

That property is the definition, and everything else follows from it. For this paper you need to know that the delta is defined by its sifting property and that calling it a function is a convenient abuse of language. A question asking "what is the Dirac delta function" wants the pulse construction, the area of one, and the sifting property.

Its transform

Apply the sifting property to the kernel, taking g(t) to be e to the minus st. The integral picks out the value of the kernel at t = a.

So the general result is e to the minus as, with no denominator.

Compare that with the unit step, whose transform is e to the minus as over s. The extra one over s in the step's transform is the integration that turns an impulse into a step, exactly as the integral theorem says.

The case a = 0, where the books disagree and you should know why

Putting a = 0 in the general result gives 1, and that is what every book on MU's reading list prints: the transform of delta(t) is 1. It is the only transform in this whole paper that does not tend to zero as s grows, which is itself a signal that the delta is not an ordinary function.

In one line

The special functions this label means are the gamma function, the error function, the sine integral and the Bessel function of order zero.

MU's label, and the problem with it

Her label is "Laplace Transformation of Special Function". She does not say which functions, and "special function" is not defined anywhere in her print. The three books on her reading list all have a chapter of that name, and the four functions below are what those chapters contain. That is the honest basis for this chapter, and it is recorded rather than assumed.

The gamma function

The gamma function extends the factorial to arguments that are not whole numbers. For a positive whole number n, gamma of n is (n minus 1) factorial.

Its two key properties:

The first is the recurrence that makes it a factorial: gamma of (n plus 1) is n times gamma of n. The second is the value that makes non-whole arguments useful, and it is the reason a square root of pi appears in transforms.

Its use here: the transform of t to the n, which the earlier chapter gave as n factorial over s to the n plus one, holds for non-whole n with the factorial replaced by gamma of (n plus 1).

Check the second against the general rule: gamma of one half is root pi, and s to the power one half is root s, so the answer is root pi over root s. It is.

The error function

The error function is the integral of the bell curve, scaled so that it tends to 1.

It has no expression in terms of elementary functions, which is exactly why it is given a name. It appears in every problem about diffusion and in every calculation with a normal distribution.

And its transform, which is the standard result:

The derivation uses the series for the error function and the gamma-function transform above, term by term.

The sine integral

The sine integral, written Si(t), is the integral of sin u over u from 0 to t. The integrand is the function whose transform the dividing-by-t chapter computed, and this is its integral.

Its transform follows from the transform of sin t over t, which is the arctangent of one over s, divided by s by the integral theorem.

That one line uses two of this module's theorems in sequence, which is why it is a satisfying examination answer: the dividing-by-t rule to get the arctangent, then the integral rule to divide by s.

In one line

The transform of a convolution is the product of the transforms, and a convolution is not a product.

The problem it solves

The linearity chapter warned that the transform of a product is not the product of the transforms. So the natural question is: if you multiply two transforms together, what have you got the transform of?

The answer is not f times g. It is a new operation on f and g called their convolution, and the theorem that says so is MU's "Convolution Theorem".

What a convolution is

The convolution of f and g, written f star g, is defined by the integral below.

Read the integrand carefully. As u runs from 0 to t, f is evaluated forwards from 0 to t and g is evaluated backwards from t to 0. The two functions slide past each other in opposite directions, and the integral adds up all the products.

That is why it is not a product: at a given time t, the convolution depends on the whole history of both functions up to t, not just on their values at t.

The theorem

So the correct statement is: a product of transforms corresponds to a convolution of functions.

What it means physically, which is why it exists

A linear system, given an impulse at time zero, produces some response. Call it g. Now feed the system an arbitrary input f.

Think of f as a great many impulses, one at each instant u, of size f(u). Each one produces a copy of the response g, starting at time u, scaled by f(u). At time t, the impulse that arrived at time u has been running for t minus u, so its contribution is f(u) times g(t minus u). Adding up all of them gives exactly the convolution integral.

So: the output of a linear system is the convolution of the input with the impulse response. That single sentence is the foundation of signal processing, and the convolution theorem is what turns it into a multiplication.

It is also why a digital filter is implemented as a convolution, and why the Fast Fourier Transform is used to do it: transform both, multiply, transform back, because multiplying is cheap and convolving is not.

Worked convolutions, from the definition

The first says the convolution of 1 with 1 is t. Check it with the theorem: the transform of 1 is one over s, so the product is one over s squared, which is the transform of t.

In one line

The proof reverses the order of a double integral, and the forward direction of the theorem computes transforms that no other rule reaches.

The proof

Write the product of the two transforms as a double integral.

F(s) is the integral over one variable, say v, of e to the minus sv times f(v). G(s) is the integral over another, say w, of e to the minus sw times g(w). Multiplying them gives a double integral over the whole quarter plane where v and w are both non-negative, of e to the minus s(v plus w) times f(v) times g(w).

Now change variables. Put t = v plus w and keep v, so w is t minus v. The Jacobian of that change is 1, so the area element is unchanged.

The region matters and is where the work is. In the (v, w) quarter plane, v and w are both at least zero. In the (v, t) variables that becomes: v at least zero, and t at least v, since w = t minus v must be non-negative. So for a given t, v runs from 0 to t, and t runs from 0 to infinity.

Reverse the order so that t is the outer variable.

The inner integral over v, from 0 to t, of f(v) times g(t minus v), is exactly the convolution. The outer integral over t of e to the minus st times that convolution is the defining integral of its transform.

So the product of the transforms is the transform of the convolution, which is the theorem.

The one step worth practising is drawing the region and reading off the new limits. Getting v from 0 to t rather than 0 to infinity is the whole of it, and a sketch of the wedge between the two lines settles it.

Using it forwards: a transform no other rule gives

The forward direction is the half students never practise, and it is set.

Find the transform of the integral below.

That is the convolution of sin t with cos t. By the theorem, its transform is the product of their transforms.

So the answer is s over (s squared plus 1) squared, with no integration at all.

And the convolution itself, if you want it, can be done by a trigonometric identity and gives t sin t over 2.

Check that against the table: the transform of t sin at is 2as over (s squared plus a squared) squared, so with a = 1 the transform of t sin t over 2 is s over (s squared plus 1) squared. The two routes agree.

In one line

The inverse transform answers the question "whose transform is this", and it is the half of the work that actually produces the answer.

The definition

If F(s) is the Laplace transform of f(t), then f(t) is the inverse Laplace transform of F(s), written with an inverse superscript on the L.

That is all the definition says. There is no new integral to learn for this paper: the inverse is found by recognising F as a transform you already know, not by computing anything.

There is a formula for the inverse as a contour integral in the complex plane, called the Bromwich integral, and it is beyond this syllabus. MU does not name it and no question needs it.

Why the question has only one answer

Because of the uniqueness result quoted in the existence chapter: two continuous functions with the same transform are the same function.

So if you find any f whose transform is the F in front of you, it is the answer. You do not have to worry that some other function has the same transform, and you do not have to justify your method. Guessing, checking, and stopping is a complete and rigorous procedure.

That licence is worth knowing, because it means the whole of the inverse-transform technique is a set of recognition patterns rather than a calculation.

Linearity, again

The inverse is linear, for the same reason the forward transform is.

So the plan for any F is always: break it into pieces each of which is a table row, and invert each piece. Every chapter in this section is a technique for doing that breaking.

The five techniques, and where each applies

That table is the whole of this section, and reading it is how you decide what to do with an F you have never seen.

The first thing to do with any F

Look at the denominator, and in this order.

1. Does it factorise into linear factors? Then partial fractions, and the answer is exponentials.
2. Is it an irreducible quadratic? Then complete the square, and the answer oscillates.
3. Is there a factor of s on its own? Then there is a constant term in the answer, or an integration.
4. Is there an exponential in s anywhere? Then something is delayed, and there will be a unit step in the answer.
5. Is a factor raised to a power? Then there will be a power of t multiplying something.

In one line

Make F(s) look like a row of the table, by supplying missing constants and normalising the coefficient of s.

The table, inverted

Note the divisors in the right column. They are there because the forward table has a factorial or an a in the numerator, so inverting a bare 1 leaves that factor to be divided out. Those divisors are where most inverse marks are lost.

The three adjustments

One: supply a missing constant. The sine row needs an a on top. If the numerator is 1, write it as one over a times a.

The second line is the one that catches people: a squared is 2, so a is root 2, and the answer has root 2 in it. There is no rule that the a must be a whole number.

Two: normalise the coefficient of s. The table always has a leading coefficient of 1 on the s. If yours is not 1, divide top and bottom.

For the third: divide top and bottom by 4 to get one quarter over (s squared plus nine quarters), so a is three halves, and the answer is one quarter times sin(3t/2) divided by three halves, which is sin(3t/2) over 6.

Three: split the numerator. A numerator with several terms goes into several table rows.

The last line is worth a look: instead of splitting into a cosh and a sinh, notice that the fraction cancels to one over (s minus 1), because the numerator is a factor of the denominator. Always try to cancel before you split.

Reading a power of s in the denominator

For the last: split it into one over s squared plus 2 over s cubed, which inverts to t plus 2 times t squared over 2, which is t plus t squared.

Reading a shifted denominator

Any F in which every s appears as (s minus a) has an e to the at in the answer, which is the first shifting theorem read backwards.

In one line

If the quadratic denominator does not factorise, complete the square, and the shift comes out as an exponential.

When to reach for it

Look at the discriminant of the quadratic in the denominator. If it is negative, the quadratic does not factorise over the reals, partial fractions cannot help, and completing the square is the method.

Checking the discriminant first takes ten seconds and saves the two minutes spent trying to factorise something that will not.

Completing the square

For s squared plus bs plus c, take half of b, square it, add and subtract.

In each case the result is (s minus a) squared plus b squared, and the two table rows that fit are the shifted sine and the shifted cosine.

The two rows

Note what a and b do. a comes out as the exponential and b as the frequency, and a's sign is the sign inside the bracket reversed: (s plus 3) squared means a is minus 3, so the answer decays.

Worked, numerator constant

Each is the same three steps: complete the square, read a and b, divide by b.

Worked, numerator containing s

This is the case that needs the extra step, and it is the commoner one in practice, because a second-order differential equation nearly always produces it.

Invert the function below.

Complete the square: the denominator is (s plus 2) squared plus 4, so a is minus 2 and b is 2.

Now the numerator must be written in terms of s plus 2, because that is what the cosine row needs on top. Write s plus 5 as (s plus 2) plus 3.

Now split into two rows.

In one line

Split the fraction into one term per factor, find the constants, and invert each term as an exponential.

MU's label

"Partial fractions Methods". The method is from school, and it is worth doing from the beginning, because a student who half remembers it will lose marks here rather than in the transform.

The shape

If the denominator factorises into distinct linear factors, write one fraction per factor with an unknown constant on top.

Then find A and B, and invert each term with the exponential row.

Finding the constants: the two methods

Method one: multiply up and compare coefficients. Multiply both sides by the whole denominator. That gives an identity between polynomials, and comparing the coefficients of each power of s gives simultaneous equations.

For the example above, multiplying up gives 1 equal to A(s minus 2) plus B(s minus 1). Comparing the s terms: 0 equals A plus B. Comparing the constants: 1 equals minus 2A minus B. Solving: A is minus 1 and B is 1.

Method two: substitute the roots. Multiply up as before, then put s equal to each root in turn. Each substitution kills all but one unknown.

Putting s = 1: 1 equals A(1 minus 2), so A is minus 1. Putting s = 2: 1 equals B(2 minus 1), so B is 1.

Method two is faster and should be your default. It is the same idea as the cover-up rule, which has its own chapter.

So:

And the inverse follows at once.

Worked, three factors

By substitution: at s = 0 the equation is 1 equals A times 1 times 2, so A is one half. At s = minus 1 it is 1 equals B times minus 1 times 1, so B is minus 1. At s = minus 2 it is 1 equals C times minus 2 times minus 1, so C is one half.

Worked, with a numerator

A numerator changes nothing about the method, only the arithmetic.

At s = minus 1: 2(minus 1) plus 3 is 1, and the other factor is 1, so A is 1. At s = minus 2: 2(minus 2) plus 3 is minus 1, and the other factor is minus 1, so B is 1.

In one line

A factor repeated n times needs n terms, one for each power, and the answer carries powers of t.

The shape

If a linear factor appears n times, you need a term for every power from 1 to n.

Writing only the highest power, C over (s minus 2) squared, is the standard mistake, and it makes the identity impossible to satisfy.

Why every power is needed

Count the unknowns against the equations. Multiplying up gives an identity between polynomials of degree 2, which has three coefficients, so three equations. Three equations need three unknowns, and the three terms above supply exactly three.

Leave out the middle term and you have two unknowns for three equations, which in general has no solution. That is the algebraic reason, and it is the answer to "why do we need the lower powers".

Finding the constants

Substitution still works for the repeated root's highest power and for every distinct root, but not for the lower powers of the repeated factor, because substituting the repeated root kills them too.

So the routine is:

1. Substitute each distinct root to get its constant.
2. Substitute the repeated root to get the constant on the highest power.
3. Get the remaining constants by comparing coefficients, or by substituting any convenient extra value of s.

Worked, on the example above. Multiplying up: 1 equals A(s minus 2) squared plus B(s minus 1)(s minus 2) plus C(s minus 1).

At s = 1: 1 equals A times 1, so A is 1. At s = 2: 1 equals C times 1, so C is 1. Now put s = 0: 1 equals A times 4 plus B times 2 plus C times minus 1, that is 1 equals 4 plus 2B minus 1, so 2B is minus 2 and B is minus 1.

Inverting, and where the t comes from

The row for a repeated factor carries a power of t.

So the example above inverts as follows.

A repeated factor means a t in the answer. That is the single fact to carry away, and it has a physical meaning: a repeated root is the boundary between oscillation and pure decay, and the t is the system on that boundary.

Worked, a repeated factor at zero

In one line

An irreducible quadratic factor needs a linear numerator, As plus B, and it produces a sine or a cosine.

The shape

If the denominator has a quadratic factor that does not factorise, the term over it needs two unknowns, not one.

The reason is the same counting argument as for repeated factors. Multiplying up gives a polynomial identity of degree 2, so three coefficients and three equations, and three unknowns are needed.

Putting a single constant over the quadratic gives two unknowns for three equations, and it generally has no solution.

Finding the constants

The root substitution works for the linear factors. For the quadratic's two unknowns, compare coefficients or substitute extra values.

Worked, on the example. Multiplying up: 1 equals A(s squared plus 4) plus (Bs plus C)(s plus 1).

At s = minus 1: 1 equals A times 5, so A is one fifth.

Comparing the coefficients of s squared: 0 equals A plus B, so B is minus one fifth.

Comparing the constants: 1 equals 4A plus C, so C is 1 minus four fifths, which is one fifth.

Inverting it

Split the quadratic's term into its s part and its constant part, and use the cosine row and the sine row.

The sin over 10 is the one fifth from C, divided by the b of 2 that the sine row needs.

An irreducible quadratic means an oscillation in the answer. That is the fact to carry, and it pairs with the two from the previous chapters: a distinct linear factor gives an exponential, a repeated one gives a t times an exponential, and an irreducible quadratic gives a sine and a cosine.

Worked, a quadratic with a linear term in it

If the quadratic itself has a middle term, complete the square after splitting.

At s = 0: 1 equals A times 2, so A is one half. Comparing s squared coefficients: 0 equals A plus B, so B is minus one half. Comparing constants: 1 equals 2A, consistent; comparing s coefficients: 0 equals 2A plus C, so C is minus 1.

Now the second term needs the completing-the-square treatment. The denominator is (s plus 1) squared plus 1, and the numerator, minus s over 2 minus 1, must be written in terms of s plus 1: it is minus (s plus 1)/2 minus 1/2.

In one line

To find the constant over a distinct linear factor, cover that factor up and put its root into everything else.

The rule

For a distinct linear factor (s minus a), the constant on top of A over (s minus a) is found by deleting that factor from F(s) and evaluating what is left at s = a.

Nothing else. No multiplying up, no simultaneous equations, no comparing coefficients.

Worked

Take the fraction below.

For the constant over (s minus 1): cover up (s minus 1), leaving one over (s minus 2)(s minus 3), and put s = 1. That gives one over (minus 1)(minus 2), which is one half.

For (s minus 2): cover it, leaving one over (s minus 1)(s minus 3), and put s = 2. That gives one over (1)(minus 1), which is minus 1.

For (s minus 3): cover it, leaving one over (s minus 1)(s minus 2), and put s = 3. That gives one over (2)(1), which is one half.

Three constants in about fifteen seconds. Doing the same by comparing coefficients means three simultaneous equations in three unknowns.

Why it works

Multiply F by (s minus a). On the right-hand side, that clears the denominator of the term you want and leaves (s minus a) as a factor on every other term.

Now let s tend to a. Every other term has a factor going to zero, so it vanishes, and only A is left. On the left, (s minus a) times F is F with the factor cancelled, evaluated at a.

So the rule is a limit, and taking the limit is what the covering up is doing.

Those two lines are the same limit written two ways, and they agree, which is the rule in miniature.

The two conditions

The factor must be linear. For a quadratic factor the numerator has two unknowns and one substitution cannot find both.

The factor must not be repeated. For a repeated factor, covering up one copy still leaves another in the denominator, and the substitution divides by zero. The rule then gives only the constant on the highest power, and the lower ones need the derivative method of the repeated-factors chapter.

The highest power of a repeated factor

The rule does work for that one case, which is worth stating separately.